Inverses of Exponential Functions
Introduced| English | Meaning |
|---|---|
| logarithmic function/ˌlɒɡəˈrɪθmɪk ˈfʌŋkʃn/ | A function giving the exponent to which a fixed positive base other than one must be raised to obtain its positive input. |
| inverse/ɪnˈvɜːs/ | A function that reverses another function's mapping on a suitable one-to-one domain. |
| domain/dəˈmeɪn/ | The set of input values for which a function is defined. |
| range/reɪndʒ/ | The set of output values actually attained by a function on its stated domain. |
| vertical asymptote/ˈvɜːtɪkl ˈæsɪmptəʊt/ | A vertical line approached by a curve as its output grows without bound near a particular input. |
Recover time from a growth factor
- A quantity doubles each hour. Its growth factor after $t$ hours is $E(t)=2^t$, with $E(0)=1$. A factor of 8 corresponds to 3 hours because $2^3=8$.
- The logarithmic function 对数函数 $L(v)=\log_2 v$ is the inverse 反函数: it recovers the time from the growth factor. The abstract functions extend beyond this example; a context starting at time zero uses only $t\ge0$ and factors $v\ge1$.
Pair functions with a valid matching base
- For $b>0$ and $b\ne1$, the functions $E(u)=b^u$ and $L(v)=\log_b v$ undo each other. The output of either becomes the input of the other.
- The bases must match. For example, $\log_2(2^3)=3$, but $\log_{10}(2^3)=\log_{10}8\approx0.903$, so a common logarithm does not undo a base-2 exponential directly.
The logarithmic function $\log_b x$ is the ____ of the exponential function $b^x$.
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They undo each other, so the log function is the inverse of the exponential with the same base.
Reverse valid points
- Since $2^3=8$, the point $(3,8)$ on $E$ becomes $(8,3)$ on $L$. Similarly, $2^{-1}=1/2$ gives the inverse point $(1/2,-1)$.
- Every pair $(s,t)$ becomes $(t,s)$, a reflection across $y=x$. Use equal scales on the two axes when drawing the reflection. Check that the original point is on the function before swapping its coordinates: $(2,8)$ is not on $y=2^x$.
What is the output of $E(u)=2^u$ at input 2?
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2²=4, so (2,4) is on the graph. The point (2,8) is not.
Since $2^{-1}=1/2$, which point lies on $L(v)=\log_2 v$?
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Exchange the input −1 and output 1/2: log₂(1/2)=−1.
Exchange domain, range and asymptotes
- The exponential has domain 定义域 all real numbers and range 值域 $(0,\infty)$. Its inverse has domain $(0,\infty)$ and range all real numbers. Thus an exponential can output 1, but cannot output zero or a negative value.
- The exponential's horizontal asymptote $y=0$ becomes the logarithm's vertical asymptote 竖直渐近线 $x=0$. The intercepts also exchange: $(0,1)$ on $E$ becomes $(1,0)$ on $L$.
The exponential has a horizontal asymptote; its inverse, the log, has a ____ asymptote instead.
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Reflecting across $y=x$ turns the horizontal asymptote $y=0$ into the vertical asymptote $x=0$.
The domain of $\log_b x$ is $x > 0$, which was the range of the exponential.
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Inverses swap domain and range: the exponential's range ($y>0$) becomes the log's domain ($x>0$).
Include bases between zero and one
- If $b>1$, both $b^x$ and $\log_b x$ are increasing. If $0<b<1$, both are decreasing. Reflection exchanges input and output; it does not turn a decreasing function into an increasing one.
- For $b=1/2$, $E(3)=1/8$ and $L(1/8)=3$. Here the exponential approaches $y=0$ as $x$ increases, while the logarithm increases without bound as its input approaches zero from the positive side.
Evaluate $\log_{1/2}(1/8)$.
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(1/2)³=1/8, so the inverse returns 3. Both basic functions are decreasing.
State the domain of each identity
- The identity $\log_b(b^u)=u$ holds for every real $u$, since $b^u$ is always positive. For example, $\log_2(2^7)=7$.
- The reverse identity $b^{\log_b v}=v$ requires $v>0$, since the logarithm must be defined first. Thus $10^{\log_{10}5}=5$ is valid, while $2^{\log_2(-5)}$ is not a real expression.
An algebraic cancellation does not remove the original input restriction. For $b^{\log_b v}$, check $v>0$ before applying the identity.
What is $\log_2\!\left(2^5\right)$?
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The log undoes the exponential: $\log_2(2^5) = 5$. Inverse functions cancel.
Select all true statements about the exponential and its inverse.
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They are inverses, not the same function. The other three correctly describe the relationship.
The identity $2^{\log_2 v}=v$ is valid for every real v, including negative values.
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The inner logarithm must be defined first, so v>0 is required.
Exchange additive and multiplicative changes
- For $E(u)=2^u$, increasing the input by 1 multiplies the output by 2: inputs $0,1,2,3$ give outputs $1,2,4,8$.
- For $L(v)=\log_2 v$, multiplying the input by 2 adds 1 to the output: inputs $1,2,4,8$ give outputs $0,1,2,3$. More generally, multiplying a positive input by $b^h$ adds $h$ to its logarithm.
For $L(v)=\log_2 v$, find $L(8)-L(4)$.
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L(8)=3 and L(4)=2. Doubling the input adds 1 to the output.
Include a scale factor in the logarithm
- The form $L(v)=a\log_b v$ requires $a\ne0$, $b>0$ and $b\ne1$. It still has positive inputs and all real outputs. If $a=0$, it is constant and has no inverse function.
- Solve $u=a\log_b v$ by dividing by $a$, then exponentiating: the inverse is $E(u)=b^{u/a}$. For $L(v)=2\log_2 v$, the inverse is $E(u)=2^{u/2}$, not $2\cdot2^u$. This inverse also has $E(0)=1$.
What is the inverse of $L(v)=2\log_2 v$?
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From u=2log₂v, divide by 2 to get u/2=log₂v. Then v=2^(u/2).
Track the scaled changes and signs
- For $L(v)=2\log_2 v$, doubling the input adds 2 to the output: $L(1)=0$, $L(2)=2$, $L(4)=4$ and $L(8)=6$. To increase the output by only 1, multiply the input by $2^{1/2}=\sqrt2$ instead.
- In general, multiplying the input by $b$ changes the output by $a$. A negative $a$ reverses the direction of the basic logarithm: $-\log_2 v$ decreases, while $-\log_{1/2}v$ increases. The inverse must use the same scale factor in its exponent denominator.
Reflect a matching exponential and logarithm
Start with (3,8) and (8,3). Change the matching base and logarithmic scale factor while retaining the inverse relationship.
For $L(v)=2\log_2 v$, find $L(8)$.
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log₂8=3, then multiply by 2. The matching inverse sends 6 back to 8.
Which statement correctly describes these functions on positive inputs?
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For base 2 the basic logarithm increases, while for base 1/2 it decreases. Multiplying by −1 reverses each direction.
Use the matching inverse to solve an equation
- For $10^x=1000$, apply $\log_{10}$ to both sides: $x=\log_{10}1000=3$. Substitution checks that $10^3=1000$.
- For $2\log_2 v=6$, divide by 2 to get $\log_2 v=3$, then use $v=2^3=8$. The result satisfies the required domain $v>0$.
Match the base and scale factor, exchange valid input-output pairs, and keep the two composition domains. Exponential outputs change multiplicatively as inputs change additively; logarithmic inputs change multiplicatively as outputs change additively.
Explain a reflected pair
- The point $(3,8)$ lies on $E(u)=2^u$. Find its matching point on $L(v)=\log_2 v$ and check it with a logarithmic equation.
- State the domain of $L$ and explain why reflecting across $y=x$ exchanges the coordinates. Then find the inverse of $L(v)=2\log_2 v$.
The point $(3,8)$ lies on $E(u)=2^u$. Find the corresponding point on $L(v)=\log_2 v$, check it, and state the domain of L. Then find the inverse of $L(v)=2\log_2 v$.
2³=8 and log₂8=3. For the scaled logarithm, divide by 2 before exponentiating.
The point $(3,8)$ lies on $y=2^x$. Which corresponding point lies on its inverse $y=\log_2 x$?
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The original relation is 2³=8. Its inverse relation is log₂8=3, so the coordinates swap to (8,3).