H Algebra 2/Trig · Unit 7
Switch x and y, then solve. An exponential undoes with a log. A log undoes with an exponential.
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Switch the variables
Replace f(x) with y. Switch x and y. Then solve for y.
The new equation is the inverse.
Exponential inverses are logs
For y = a·bx + c, subtract c, divide by a, then write a log.
A natural exponential undoes with ln.
Log inverses are exponentials
For y = a logb(x − h) + k, subtract k, divide by a, then write a power.
The base of the power is the base of the log.
No coefficient other than 1
f(x) = 4x + 2
Replace f(x) with y.
y = 4x + 2
Switch x and y.
x = 4y + 2
Subtract 2 from both sides.
x − 2 = 4y
Rewrite in log form. The log answer is the exponent.
\(f^{-1}(x) = \log_{4}\left(x - 2\right)\)
Answer: \(f^{-1}(x) = \log_{4}\left(x - 2\right)\)
f(x) = 2x − 5
Replace f(x) with y.
y = 2x − 5
Switch x and y.
x = 2y − 5
Add 5 to both sides.
x + 5 = 2y
Rewrite in log form. The log answer is the exponent.
\(f^{-1}(x) = \log_{2}\left(x + 5\right)\)
Answer: \(f^{-1}(x) = \log_{2}\left(x + 5\right)\)
f(x) = 5x + 1
Replace f(x) with y.
y = 5x + 1
Switch x and y.
x = 5y + 1
Subtract 1 from both sides.
x − 1 = 5y
Rewrite in log form. The log answer is the exponent.
\(f^{-1}(x) = \log_{5}\left(x - 1\right)\)
Answer: \(f^{-1}(x) = \log_{5}\left(x - 1\right)\)
f(x) = 3x − 4
Replace f(x) with y.
y = 3x − 4
Switch x and y.
x = 3y − 4
Add 4 to both sides.
x + 4 = 3y
Rewrite in log form. The log answer is the exponent.
\(f^{-1}(x) = \log_{3}\left(x + 4\right)\)
Answer: \(f^{-1}(x) = \log_{3}\left(x + 4\right)\)
g(x) = log2 (x) + 3
Replace g(x) with y.
Switch x and y.
x = log2 (y) + 3
Subtract 3 from both sides.
x − 3 = log2 (y)
Rewrite in exponential form.
\(g^{-1}(x) = 2^{x - 3}\)
Answer: \(g^{-1}(x) = 2^{x - 3}\)
g(x) = log3 (x − 2)
Replace g(x) with y.
Switch x and y.
x = log3 (y − 2)
Rewrite in exponential form.
\(g^{-1}(x) = 3^{x} + 2\)
Answer: \(g^{-1}(x) = 3^{x} + 2\)
A coefficient in front
f(x) = 3(2x) − 6
Replace f(x) with y.
y = 3(2x) − 6
Switch x and y.
x = 3(2y) − 6
Add 6 to both sides.
x + 6 = 3(2y)
Divide by 3.
x + 63 = 2y
Rewrite in log form. The log answer is the exponent.
\(f^{-1}(x) = \log_{2}\left(\dfrac{x + 6}{3}\right)\)
Answer: \(f^{-1}(x) = \log_{2}\left(\dfrac{x + 6}{3}\right)\)
f(x) = 2(5x) + 4
Replace f(x) with y.
y = 2(5x) + 4
Switch x and y.
x = 2(5y) + 4
Subtract 4 from both sides.
x − 4 = 2(5y)
Divide by 2.
x − 42 = 5y
Rewrite in log form. The log answer is the exponent.
\(f^{-1}(x) = \log_{5}\left(\dfrac{x - 4}{2}\right)\)
Answer: \(f^{-1}(x) = \log_{5}\left(\dfrac{x - 4}{2}\right)\)
f(x) = 4(3x) − 1
Replace f(x) with y.
y = 4(3x) − 1
Switch x and y.
x = 4(3y) − 1
Add 1 to both sides.
x + 1 = 4(3y)
Divide by 4.
x + 14 = 3y
Rewrite in log form. The log answer is the exponent.
\(f^{-1}(x) = \log_{3}\left(\dfrac{x + 1}{4}\right)\)
Answer: \(f^{-1}(x) = \log_{3}\left(\dfrac{x + 1}{4}\right)\)
f(x) = 6(2x) + 3
Replace f(x) with y.
y = 6(2x) + 3
Switch x and y.
x = 6(2y) + 3
Subtract 3 from both sides.
x − 3 = 6(2y)
Divide by 6.
x − 36 = 2y
Rewrite in log form. The log answer is the exponent.
\(f^{-1}(x) = \log_{2}\left(\dfrac{x - 3}{6}\right)\)
Answer: \(f^{-1}(x) = \log_{2}\left(\dfrac{x - 3}{6}\right)\)
g(x) = 2 · log5 (x + 1) − 4
Replace g(x) with y.
Switch x and y.
x = 2 · log5 (y + 1) − 4
Add 4 to both sides.
x + 4 = 2 · log5 (y + 1)
Divide by 2.
x + 42 = log5 (y + 1)
Rewrite in exponential form.
\(g^{-1}(x) = 5^{\dfrac{x + 4}{2}} - 1\)
Answer: \(g^{-1}(x) = 5^{\dfrac{x + 4}{2}} - 1\)
g(x) = 3 · log (x − 2) + 6
Replace g(x) with y.
Switch x and y.
x = 3 · log (y − 2) + 6
Subtract 6 from both sides.
x − 6 = 3 · log (y − 2)
Divide by 3.
x − 63 = log (y − 2)
Rewrite in exponential form.
\(g^{-1}(x) = 10^{\dfrac{x - 6}{3}} + 2\)
Answer: \(g^{-1}(x) = 10^{\dfrac{x - 6}{3}} + 2\)
A negative coefficient, a natural log, or a shift
f(x) = 2(ex) − 8
Replace f(x) with y.
y = 2(ex) − 8
Switch x and y.
x = 2(ey) − 8
Add 8 to both sides.
x + 8 = 2(ey)
Divide by 2.
x + 82 = ey
Rewrite in log form. The log answer is the exponent.
\(f^{-1}(x) = \ln\left(\dfrac{x + 8}{2}\right)\)
Answer: \(f^{-1}(x) = \ln\left(\dfrac{x + 8}{2}\right)\)
f(x) = 5(ex) + 3
Replace f(x) with y.
y = 5(ex) + 3
Switch x and y.
x = 5(ey) + 3
Subtract 3 from both sides.
x − 3 = 5(ey)
Divide by 5.
x − 35 = ey
Rewrite in log form. The log answer is the exponent.
\(f^{-1}(x) = \ln\left(\dfrac{x - 3}{5}\right)\)
Answer: \(f^{-1}(x) = \ln\left(\dfrac{x - 3}{5}\right)\)
g(x) = 4 · ln(x + 2) − 8
Replace g(x) with y.
Switch x and y.
x = 4 · ln(y + 2) − 8
Add 8 to both sides.
x + 8 = 4 · ln(y + 2)
Divide by 4.
x + 84 = ln(y + 2)
Rewrite in exponential form.
\(g^{-1}(x) = e^{\dfrac{x + 8}{4}} - 2\)
Answer: \(g^{-1}(x) = e^{\dfrac{x + 8}{4}} - 2\)
f(x) = −2(3x) + 10
Replace f(x) with y.
y = −2(3x) + 10
Switch x and y.
x = −2(3y) + 10
Subtract 10 from both sides.
x − 10 = −2(3y)
Divide by −2.
x − 10−2 = 3y
Move the negative sign into the numerator.
10 − x2 = 3y
Rewrite in log form. The log answer is the exponent.
\(f^{-1}(x) = \log_{3}\left(\dfrac{10 - x}{2}\right)\)
Answer: \(f^{-1}(x) = \log_{3}\left(\dfrac{10 - x}{2}\right)\)
g(x) = −3 · log2 (x − 4) + 6
Replace g(x) with y.
Switch x and y.
x = −3 · log2 (y − 4) + 6
Subtract 6 from both sides.
x − 6 = −3 · log2 (y − 4)
Divide by −3.
x − 6−3 = log2 (y − 4)
Move the negative sign into the numerator.
6 − x3 = log2 (y − 4)
Rewrite in exponential form.
\(g^{-1}(x) = 2^{\dfrac{6 - x}{3}} + 4\)
Answer: \(g^{-1}(x) = 2^{\dfrac{6 - x}{3}} + 4\)
f(x) = 4(6x − 1) + 2
Replace f(x) with y.
y = 4(6x − 1) + 2
Switch x and y.
x = 4(6y − 1) + 2
Subtract 2 from both sides.
x − 2 = 4(6y − 1)
Divide by 4.
x − 24 = 6y − 1
Rewrite in log form. The log answer is the exponent.
Add 1 to both sides.
\(f^{-1}(x) = \log_{6}\left(\dfrac{x - 2}{4}\right) + 1\)
Answer: \(f^{-1}(x) = \log_{6}\left(\dfrac{x - 2}{4}\right) + 1\)
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