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Rewrite 64 as $2^6$
Equate indices: $x+1=6$
$x=5$

","mcqOptions":[{"text":"

Rewrite 64 as $2^6$
Equate indices: $x+1=6$
$x=5$

","correct":true,"explanation":"Since $64=2^6$, both sides have the same base, so their indices are equal: $x+1=6$. Solving gives $x=5$."},{"text":"

Rewrite 64 as $2^5$
Equate indices: $x+1=5$
$x=4$

","correct":false,"explanation":"Rewriting 64 as $2^5$ is the error; $2^5=32$, not 64. The correct rewrite is $64=2^6$, giving $x+1=6$ and therefore $x=5$."},{"text":"

Rewrite 64 as $2^6$
Equate indices: $x=6\\text{}
$x=6$

","correct":false,"explanation":"The index on the left is $x+1$, so subtracting 1 is necessary after equating it to 6. The correct equation is $x+1=6$, which gives $x=5$."},{"text":"

Rewrite 64 as $2^6$
Equate indices: $x+1=6$
$x=7$

","correct":false,"explanation":"The base remains 2, so the exponent must be compared with 6, not treated as the final value of $x+1$. Solving $x+1=6$ gives $x=5$."}]},{"id":"58656bae-4586-4836-9ab3-61b126c32823","cardType":"basic","points":1,"front":"Convert the logarithmic form $\\log_a b = c$ into exponential form.","back":"$$a^c = b$","mcqOptions":[{"text":"$$a^c = b$","correct":true,"explanation":"By definition, $log_a b=c$ means that raising the base $a$ to the power $c$ produces $b$, so the equivalent exponential form is $a^c=b$."},{"text":"$$a^b = c$","correct":false,"explanation":"This swaps the logarithm’s argument and exponent. The definition places the base $a$ to the power of the logarithm’s value $c$, giving $a^c=b$."},{"text":"$$b^c = a$","correct":false,"explanation":"The base of the logarithm becomes the base of the exponential, not its result. Thus $a$ must be raised to $c$ to obtain $b$: $a^c=b$."},{"text":"$$c^a = b$","correct":false,"explanation":"This reverses the roles of the base and argument. The defining relationship is $a^c=b$, because $c$ is the exponent needed to produce the argument $b$."}]},{"id":"88db4d11-0b6c-4d4a-b74e-36c27aa92076","cardType":"cloze","points":null,"front":"The natural logarithm, written as $\\ln x$, is a logarithm with a base of $\\text{[...]}$ (approximately 2.718).","back":"The natural logarithm, written as $\\ln x$, is a logarithm with a base of $\\boxed{e}$ (approximately 2.718).","mcqOptions":[{"text":"$$e$","correct":true,"explanation":"The natural logarithm is defined as the logarithm to base $e$, where $e$ is approximately 2.718. Therefore, $ln x$ means $log_e x$."},{"text":"$$2$","correct":false,"explanation":"Base 2 gives the binary logarithm, not the natural logarithm. The notation $ln x$ specifically means logarithm to base $e$, with $e$ approximately 2.718."},{"text":"$$10$","correct":false,"explanation":"Base 10 is the common logarithm, usually written $log x$, rather than the natural logarithm. Natural logarithms use the base $e$."},{"text":"$$pi$","correct":false,"explanation":"$$pi$ is an important mathematical constant, but it is not the base used by natural logarithms. The natural logarithm has base $e$, approximately 2.718."}]},{"id":"ca8133c6-4501-44fd-b12d-b8d247ca64ba","cardType":"basic","points":1,"front":"How are the graphs of $y=e^x$ and $y=\\ln x$ geometrically related?","back":"They are inverses, meaning they reflect in the line $y=x$.","mcqOptions":[{"text":"They are inverses, meaning they reflect in the line $y=x$.","correct":true,"explanation":"The functions $y=e^x$ and $y=ln x$ undo each other, so they are inverse functions. Inverse graphs reflect across the line $y=x$."},{"text":"They are reflections, meaning they reflect in the $x$-axis.","correct":false,"explanation":"Reflection in the $x$-axis would change $y=e^x$ into $y=-e^x$, not into $ln x$. The two given graphs reflect in the line $y=x$ because they are inverses."},{"text":"They are translations, meaning one is shifted from the other.","correct":false,"explanation":"The graphs are not translations of one another; their domains and shapes differ. They are inverse functions, so the correct geometric relationship is reflection in $y=x$."},{"text":"They are reflections, meaning they reflect in the $y$-axis.","correct":false,"explanation":"Reflection in the $y$-axis would replace $x$ by $-x$, which does not produce $ln x$ from $e^x$. Inverse functions reflect in $y=x$."}]},{"id":"8d8333f8-e6d4-40c3-bfc9-f28aeae34c61","cardType":"basic","points":1,"front":"Express $\\log_a M + \\log_a N$ as a single logarithm.","back":"$$\\log_a(MN)$","mcqOptions":[{"text":"$$\\log_a(MN)$","correct":true,"explanation":"The logarithm product law states that adding logarithms with the same base combines their arguments by multiplication: $log_a M+log_a N=log_a(MN)$."},{"text":"$$log_a\\textstylerac{M}{N}$","correct":false,"explanation":"A quotient inside the logarithm corresponds to subtraction, not addition. Addition uses the product law, so the correct single logarithm is $log_a(MN)$."},{"text":"$$log_a(M+N)$","correct":false,"explanation":"There is no logarithm law that changes a sum of logarithms into the logarithm of a sum. The arguments multiply, giving $log_a(MN)$."},{"text":"$$log_a(M-N)$","correct":false,"explanation":"A difference inside the argument is not produced by adding logarithms. The product law applies, so the result must be $log_a(MN)$."}]},{"id":"bf1eada3-54e8-411a-8bb6-6b80800d235d","cardType":"basic","points":1,"front":"Express $\\log_a M - \\log_a N$ as a single logarithm.","back":"$$\\log_a\\left(\\frac{M}{N}\\right)$","mcqOptions":[{"text":"$$\\log_a\\left(\\frac{M}{N}\\right)$","correct":true,"explanation":"The logarithm quotient law states that subtracting logarithms with the same base divides their arguments: $log_a M-log_a N=log_arac{M}{N}$."},{"text":"$$log_a(MN)$","correct":false,"explanation":"Multiplication inside the logarithm corresponds to addition, not subtraction. The quotient law gives $log_a\\textbackslashleft(\\textbackslashfrac{M}{N}\\textbackslashright)$ for the stated expression."},{"text":"$$log_a(M-N)$","correct":false,"explanation":"The subtraction must occur between the logarithms before applying the quotient law; it does not become subtraction inside one argument. The correct result is $log_a\\textbackslashleft(\\textbackslashfrac{M}{N}\\textbackslashright)$."},{"text":"$$log_a\\textbackslashleft(\\textbackslashfrac{N}{M}\\textbackslashright)$","correct":false,"explanation":"Reversing numerator and denominator changes the sign: it represents $log_a N-log_a M$. The given order produces $M/N$, not $N/M$."}]},{"id":"a9e1b0da-5e3c-4ee9-9d33-7e4d9b07cf61","cardType":"basic","points":1,"front":"Express $p \\log_a M$ as a single logarithm.","back":"$$\\log_a(M^p)$","mcqOptions":[{"text":"$$\\log_a(M^p)$","correct":true,"explanation":"The power law moves a coefficient multiplying a logarithm into the exponent of its argument: $plog_a M=log_a(M^p)$."},{"text":"$$log_a M+p$","correct":false,"explanation":"Adding $p$ outside the logarithm does not represent multiplication by $p$. The coefficient becomes an exponent, giving $log_a(M^p)$."},{"text":"$$log_a(M+p)$","correct":false,"explanation":"The coefficient is not added to the argument; it becomes an exponent. Therefore the correct transformed expression is $log_a(M^p)$."},{"text":"$$log_a(M^{1/p})$","correct":false,"explanation":"Using $1/p$ reverses the power-law transformation and would correspond to dividing the logarithm by $p$. Multiplication by $p$ gives the exponent $p$."}]},{"id":"246c146e-d806-4632-bb4c-7b70d6f25867","cardType":"cloze","points":null,"front":"The argument $x$ of any logarithm $\\log_a x$ must satisfy the condition $\\text{[...]}$.","back":"The argument $x$ of any logarithm $\\log_a x$ must satisfy the condition $\\boxed{x > 0}$.","mcqOptions":[{"text":"$$x > 0$","correct":true,"explanation":"A logarithm is defined only when its argument is positive, so $x$ must be greater than zero. The value $x = 0$ is excluded because no finite power of a valid base produces zero."},{"text":"$$x \\geq 0$","correct":false,"explanation":"Including zero is the error: logarithms cannot have an argument of zero, since no real exponent makes a positive base equal to zero. The required condition is $x > 0$."},{"text":"$$x < 0$","correct":false,"explanation":"Negative arguments are not allowed for real logarithms, because a positive base raised to a real power remains positive. The argument must instead satisfy $x > 0$."},{"text":"$$x \\neq 0$","correct":false,"explanation":"Excluding only zero is insufficient because negative arguments are also outside the domain of real logarithms. Both restrictions are captured by the correct condition $x > 0$."}]},{"id":"ef6d0601-94f2-42cd-9863-abddfdf0602f","cardType":"basic","points":2,"front":"Solve $5^x = 18$ for $x$ using natural logarithms.","back":"$$x = \\frac{\\ln 18}{\\ln 5}$","mcqOptions":[{"text":"$$x = \\frac{\\ln 18}{\\ln 5}$","correct":true,"explanation":"Taking natural logarithms gives $x\\ln 5=\\ln 18$. Dividing by $\\ln 5$ isolates $x$, producing the stated logarithmic expression."},{"text":"$$x = \\frac{\\ln 5}{\\ln 18}$","correct":false,"explanation":"This reverses the logarithmic ratio. From $x\\ln 5=\\ln 18$, division must be by $\\ln 5$, so the correct result is $x = \\frac{\\ln 18}{\\ln 5}$."},{"text":"$$x = \\ln\\left(\\frac{18}{5}\\right)$","correct":false,"explanation":"Subtracting logarithms would apply to a quotient, but the unknown is an exponent. Taking logarithms gives $x\\ln 5=\\ln 18$, hence $x = \\frac{\\ln 18}{\\ln 5}$."},{"text":"$$x = \\frac{18}{5}$","correct":false,"explanation":"Dividing the original equation treats the exponent as though it were a coefficient. Logarithms are needed to bring down the power, giving $x = \\frac{\\ln 18}{\\ln 5}$."}]},{"id":"ab3a2727-2542-4c2a-a140-0b623f41b3ae","cardType":"basic","points":1,"front":"What is the standard substitution used to solve $3e^y + 10e^{-y} = 11$?","back":"Let $u = e^y$ (where $u > 0$).","mcqOptions":[{"text":"Let $u = e^y$ (where $u > 0$).","correct":true,"explanation":"Setting $u=e^y$ means $e^{-y}=\\frac{1}{u}$, so the equation becomes $3u+\\frac{10}{u}=11$. Multiplying through by $u$ then produces a quadratic."},{"text":"Let $u = e^{-y}$ (where $u > 0$).","correct":false,"explanation":"Although this quantity is positive, it does not give the usual direct substitution for the displayed equation. The standard choice is $u=e^y$, making $e^{-y}=\\frac1u$ and producing a quadratic."},{"text":"Let $u = y$ (where $u > 0$).","correct":false,"explanation":"Replacing $y$ itself does not simplify the two reciprocal exponential terms. The useful substitution is $u=e^y$, because then $e^{-y}=\\frac1u$ and the equation becomes quadratic."},{"text":"Let $u = e^y + e^{-y}$ (where $u > 0$).","correct":false,"explanation":"The sum of the exponential terms does not convert the equation into a single quadratic in the new variable. Using $u=e^y$ links the two terms through $e^{-y}=\\frac1u$."}]},{"id":"7c9918b5-57b5-45e7-9d57-b7de487d5e0e","cardType":"cloze","points":null,"front":"In the growth model $P = Ae^{kt}$, the population grows if $\\text{[...]}$ and decays if $k < 0$.","back":"In the growth model $P = Ae^{kt}$, the population grows if $\\boxed{k > 0}$ and decays if $k < 0$.","mcqOptions":[{"text":"$$k > 0$","correct":true,"explanation":"When $k>0$, the factor $e^{kt}$ increases as $t$ increases, so $P=Ae^{kt}$ grows, assuming the initial constant $A$ is positive."},{"text":"$$k < 0$","correct":false,"explanation":"A negative value makes $e^{kt}$ decrease as time increases, which represents decay. Growth requires $k>0$, while $k<0$ describes the decay case."},{"text":"$$k = 0$","correct":false,"explanation":"If $k=0$, then $e^{kt}=1$ for every time, so the population remains constant at $A$. Growth requires the positive condition $k>0$."},{"text":"$$k \\leq 0$","correct":false,"explanation":"This includes negative values, which produce decay, and zero, which produces no change. Only $k>0$ guarantees growth in the model."}]},{"id":"807ef478-51d9-40fe-bf2c-b8d2e94cda3e","cardType":"cloze","points":null,"front":"In the growth model $P = Ae^{kt}$, the population grows if $k > 0$ and decays if $\\text{[...]}$.","back":"In the growth model $P = Ae^{kt}$, the population grows if $k > 0$ and decays if $\\boxed{k < 0}$.","mcqOptions":[{"text":"$$k < 0$","correct":true,"explanation":"When $k<0$, the exponent $kt$ becomes more negative as time increases, so $e^{kt}$ decreases and the model represents decay."},{"text":"$$k > 0$","correct":false,"explanation":"A positive value makes $e^{kt}$ increase with time, giving growth rather than decay. The decay condition in this model is $k<0$."},{"text":"$$k = 0$","correct":false,"explanation":"With $k=0$, the exponential factor is always $1$, so the population stays constant instead of decreasing. Decay requires $k<0$."},{"text":"$$k \\leq 0$","correct":false,"explanation":"This condition includes zero, which gives a constant population rather than decay. A strictly negative value, $k<0$, is required for the population to decrease."}]},{"id":"08439a56-5775-4253-be44-62f766035acd","cardType":"basic","points":2,"front":"Find the derivative $\\frac{dP}{dt}$ for the exponential model $P = Ae^{kt}$.","back":"$$\\frac{dP}{dt} = kAe^{kt}$","mcqOptions":[{"text":"$$\\frac{dP}{dt} = kAe^{kt}$","correct":true,"explanation":"Using the chain rule, differentiating $e^{kt}$ gives $ke^{kt}$. Multiplying by the constant $A$ gives $\\frac{dP}{dt}=kAe^{kt}$."},{"text":"$$\\frac{dP}{dt} = Ae^{kt}$","correct":false,"explanation":"This differentiates the exponential factor but misses the chain-rule multiplier from the exponent $kt$. Differentiating $e^{kt}$ contributes $k$, giving $\\frac{dP}{dt}=kAe^{kt}$."},{"text":"$$\\frac{dP}{dt} = kAe^{k}$","correct":false,"explanation":"The variable $t$ must remain in the exponential after differentiation; only the inner function contributes the factor $k$. The derivative is $kAe^{kt}$."},{"text":"$$\\frac{dP}{dt} = -kAe^{kt}$","correct":false,"explanation":"The derivative of $e^{kt}$ has a positive sign because the exponent is $kt$, not $-kt$. Applying the chain rule gives $kAe^{kt}$."}]},{"id":"90886971-5659-4007-8060-f207a9da2316","cardType":"basic","points":1,"front":"In an exponential model $P = Ae^{kt}$, how is the initial value found?","back":"Set $t = 0$.","mcqOptions":[{"text":"Set $t = 0$.","correct":true,"explanation":"At the initial time, substitute $t=0$ into the model: $P=Ae^{k\\cdot0}=A$. Thus the constant $A$ is the initial value."},{"text":"Set $t = 1$.","correct":false,"explanation":"Substituting $t=1$ gives the population one time unit after the start, not the initial value. The initial value is found by setting $t=0$."},{"text":"Set $P = 0$.","correct":false,"explanation":"Setting the population to zero would search for a time when the model reaches zero, but an exponential model with positive $A$ never does. Use $t=0$ for the initial value."},{"text":"Set $k = 0$.","correct":false,"explanation":"Setting $k=0$ removes growth or decay and changes the model rather than selecting the initial time. The initial value is obtained by setting $t=0$."}]},{"id":"57ac20e7-27fa-4128-9545-68e381fa702d","cardType":"cloze","points":null,"front":"The range of the basic exponential function $f(x) = e^x$ is $\\text{[...]}$.","back":"The range of the basic exponential function $f(x) = e^x$ is $\\boxed{f(x) > 0}$.","mcqOptions":[{"text":"$$f(x) > 0$","correct":true,"explanation":"The exponential $e^x$ is positive for every real value of $x$. It approaches zero as $x$ tends to negative infinity but never reaches or crosses the x-axis."},{"text":"$$f(x) \\geq 0$","correct":false,"explanation":"The exponential function is never equal to zero, even though it approaches zero for very negative $x$. Therefore its range is strictly $f(x)>0$."},{"text":"$$f(x) < 0$","correct":false,"explanation":"An exponential with positive base cannot produce negative outputs. Its values are always positive, so the range is $f(x)>0$."},{"text":"$$f(x) \\geq 1$","correct":false,"explanation":"The output is at least $1$ only when $x\\geq0$. For negative $x$, $e^x$ lies between zero and one, so the full range is $f(x)>0$."}]},{"id":"462a2aad-835a-4c9c-b287-b5b5a68c6704","cardType":"basic","points":1,"front":"State the horizontal asymptote of $y = e^{x+2} - 4$.","back":"$$y = -4$","mcqOptions":[{"text":"$$y = -4$","correct":true,"explanation":"The term $e^{x+2}$ approaches zero as $x$ tends to negative infinity. Therefore the graph approaches $y=0-4$, giving the horizontal asymptote $y=-4$."},{"text":"$$y = 4$","correct":false,"explanation":"The constant is subtracted from the exponential, so the graph shifts downward by four units rather than upward. The horizontal asymptote is $y=-4$."},{"text":"$$y = 2$","correct":false,"explanation":"The $+2$ is inside the exponent, producing a horizontal shift and not the vertical asymptote. Since the exponential tends to zero, the asymptote is $y=-4$."},{"text":"$$y = 0$","correct":false,"explanation":"The exponential part has asymptote zero before the vertical translation, but subtracting four moves that line down. Hence the asymptote of the given graph is $y=-4$."}]},{"id":"b88bf444-68e5-4162-b6b1-4882f032afe1","cardType":"basic","points":2,"front":"Outline the steps to find the x-intercept of $f(x) = e^{x+2} - 4$.","back":"

Set $f(x) = 0$
$e^{x+2} = 4$
$x = \\ln 4 - 2$

","mcqOptions":[{"text":"

Set $f(x) = 0$
$e^{x+2} = 4$
$x = \\ln 4 - 2$

","correct":true,"explanation":"An x-intercept occurs where the function equals zero. Rearranging gives $e^{x+2}=4$, and taking natural logarithms produces $x+2=\\ln 4$, so $x=\\ln 4-2$."},{"text":"

Set $f(x) = 1$
$e^{x+2} = 5$
$x = \\ln 5 - 2$

","correct":false,"explanation":"The x-intercept is found by setting $f(x)=0$, not $1$. Using $f(x)=1$ leads to the wrong equation and therefore the wrong value $x=\\ln 5-2$."},{"text":"

Set $f(x) = 0$
$e^{x+2} = -4$
$x = \\ln(-4) - 2$

","correct":false,"explanation":"Adding $4$ to both sides of $e^{x+2}-4=0$ gives $e^{x+2}=4$, not $-4$. Exponentials are positive, so the correct result is $x=\\ln 4-2$."},{"text":"

Set $f(x) = 0$
$e^{x+2} = 4$
$x = 4 - 2$

","correct":false,"explanation":"After obtaining $e^{x+2}=4$, you must take a natural logarithm; the inverse of the exponential is not simply removing $e$. Thus the answer is $x=\\ln 4-2$."}]},{"id":"f9ddcde0-f2a6-4c5f-91a2-9a551b166420","cardType":"cloze","points":null,"front":"If $y = \\log_2 x$, then $\\log_4 x$ can be written in terms of $y$ as $\\text{[...]}$.","back":"If $y = \\log_2 x$, then $\\log_4 x$ can be written in terms of $y$ as $\\boxed{\\frac{y}{2}}$.","mcqOptions":[{"text":"$$\\frac{y}{2}$","correct":true,"explanation":"Since $x=2^y$, we have $\\log_4 x=\\log_{2^2}x=\\frac{1}{2}\\log_2x=\\frac{y}{2}$. Changing the base from $2$ to $4$ introduces the factor of one half."},{"text":"$$2y$","correct":false,"explanation":"Changing the base from $2$ to $4=2^2$ divides the logarithm by $2$, rather than multiplying it. Therefore $\\log_4x=\\frac{y}{2}$, not $2y$."},{"text":"$$y+2$","correct":false,"explanation":"The new base changes the logarithm multiplicatively, not by adding its exponent. Since $4=2^2$, $\\log_4x=\\frac{1}{2}\\log_2x=\\frac{y}{2}$."},{"text":"$$\\frac{2}{y}$","correct":false,"explanation":"Taking the reciprocal of $y$ confuses a logarithm with its inverse relationship. The base-change factor is determined by $4=2^2$, giving $\\log_4x=\\frac{y}{2}$."}]},{"id":"e296a454-8325-475c-8d88-cd4b814e141b","cardType":"basic","points":1,"front":"Why is the statement $\\log_a(M+N) = \\log_a M + \\log_a N$ incorrect?","back":"Log laws apply to multiplication, not addition, inside the logarithm.","mcqOptions":[{"text":"Log laws apply to multiplication, not addition, inside the logarithm.","correct":true,"explanation":"The product law is $\\log_a(MN)=\\log_aM+\\log_aN$. Addition inside the logarithm does not split into separate logarithms, so $\\log_a(M+N)$ cannot be simplified that way."},{"text":"Log laws apply to addition, not multiplication, inside the logarithm.","correct":false,"explanation":"This reverses the key logarithm rule. Multiplication splits as $\\log_a(MN)=\\log_aM+\\log_aN$, whereas addition inside the logarithm cannot be separated; the correct principle concerns multiplication."},{"text":"Log laws apply to subtraction, not addition, inside the logarithm.","correct":false,"explanation":"Subtraction is not the operation that produces a sum of logarithms. The valid product law applies to multiplication, so the correct statement says laws apply to multiplication, not addition."},{"text":"Log laws apply to division, not addition, inside the logarithm.","correct":false,"explanation":"Division gives a difference, $\\log_a(M/N)=\\log_aM-\\log_aN$, rather than the sum shown in the statement. The relevant reason is that logarithm addition comes from multiplication."}]},{"id":"51d2ed13-8156-43ef-9079-266aa8b7a11b","cardType":"basic","points":2,"front":"Find the gradient function $f'(x)$ for $f(x) = e^{x+2} - 4$.","back":"$$f'(x) = e^{x+2}$","mcqOptions":[{"text":"$$f'(x) = e^{x+2}$","correct":true,"explanation":"Using the chain rule, the derivative of $e^{x+2}$ is $e^{x+2}$ multiplied by the derivative of $x+2$, which is $1$. The constant $-4$ differentiates to zero."},{"text":"$$f'(x) = e^{x+2} - 4$","correct":false,"explanation":"This repeats the original function instead of differentiating it. The constant $-4$ disappears under differentiation, leaving $f'(x)=e^{x+2}$."},{"text":"$$f'(x) = (x+2)e^{x+1}$","correct":false,"explanation":"This incorrectly applies a power-rule-style change to the exponential. The derivative of $e^{u}$ is $e^u u'$, and here $u'=1$, so $f'(x)=e^{x+2}$."},{"text":"$$f'(x) = e^{x+2} + 4$","correct":false,"explanation":"The constant term does not become positive when differentiated; its derivative is zero. Therefore the gradient function remains $f'(x)=e^{x+2}$."}]},{"id":"6a3ea351-8ac5-4094-b2a1-0a0636aa7474","cardType":"cloze","points":null,"front":"To solve $\\ln(3x-2) = 1$, first remove the natural log by writing $\\text{[...]}$.","back":"To solve $\\ln(3x-2) = 1$, first remove the natural log by writing $\\boxed{3x - 2 = e^1}$.","mcqOptions":[{"text":"$$3x - 2 = e^1$","correct":true,"explanation":"Applying the exponential function to both sides reverses the natural logarithm: if $\\ln(3x-2)=1$, then $3x-2=e^1$. The logarithm argument must also be positive."},{"text":"$$3x - 2 = 1^e$","correct":false,"explanation":"The inverse of $\\ln$ is the exponential function with base $e$, not raising $1$ to the power $e$. Exponentiating both sides gives the correct equation $3x-2=e^1$."},{"text":"$$3x - 2 = \\ln e^1$","correct":false,"explanation":"Writing $\\ln e^1$ does not remove the logarithm and simply evaluates back to $1$. Applying the inverse operation requires exponentiating, giving $3x-2=e^1$."},{"text":"$$3x - 2 = e^{-1}$","correct":false,"explanation":"The negative exponent changes the right-hand side to the reciprocal of $e$, which would correspond to a logarithm value of $-1$. Since the given value is $1$, the equation is $3x-2=e^1$."}]}]}],"examCountdown":{"examDate":"2027-05-13","paperLabel":"Paper 1"},"heading":"Exponentials and Logarithms","title":"Exponentials and Logarithms Flashcards","topbar":{"backHref":"/as-level/maths/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"b381dbdf-209e-44e8-95f7-830172c37463","walled":"$L38"}],"$L39","$L3a"]