Logarithms and Inverse Functions

25 min
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Undoing a function

The inverse f1f^{-1} reverses what ff does: you swap xx and yy and make yy the subject.

Linear. For f(x)=2x6f(x) = 2x - 6 write y=2x6y = 2x - 6, swap: x=2y6x = 2y - 6, solve: y=x+62y = \dfrac{x+6}{2}. So f1(x)=x+62f^{-1}(x) = \dfrac{x+6}{2}.

Parabola. For f(x)=2x2f(x) = 2x^2 swapping gives x=2y2x = 2y^2, so y=±x2y = \pm\sqrt{\dfrac{x}{2}} — not a function. Restricting the domain to x0x \ge 0 gives the single branch f1(x)=x2f^{-1}(x) = \sqrt{\dfrac{x}{2}}.

Exponential. For f(x)=bxf(x) = b^x (with b>0b > 0, b1b \neq 1) swapping gives x=byx = b^y, and the value of yy is defined to be the logarithm:

y=logbx    by=xy = \log_b x \iff b^{\,y} = x

So the inverse of y=3xy = 3^x is y=log3xy = \log_3 x.

Log laws

logb(MN)=logbM+logbN,logb ⁣(MN)=logbMlogbN,logb(Mk)=klogbM\log_b(MN) = \log_b M + \log_b N, \qquad \log_b\!\left(\frac{M}{N}\right) = \log_b M - \log_b N, \qquad \log_b(M^k) = k\log_b M

and the change of base logbx=logxlogb\log_b x = \dfrac{\log x}{\log b}, which is what lets your calculator handle any base.

Worked example

A prepaid electricity meter's credit grows in a simulation as 3x3^x. Solve 3x=203^x = 20:

x=log320=log20log3=1,301030,477122,73x = \log_3 20 = \frac{\log 20}{\log 3} = \frac{1{,}30103}{0{,}47712} \approx 2{,}73

Check: 32,73203^{2{,}73} \approx 20.

The number of households connected to a new solar micro-grid is modelled by f(x)=3xf(x) = 3^x, where xx is the number of years. Which function is f1f^{-1}?

A dam's algae population doubles every day and is modelled by P=2xP = 2^x times its starting size. Solve for xx if 2x=502^x = 50. Give your answer correct to two decimal places.

A taxi fare is f(x)=2x6f(x) = 2x - 6 rand for a trip of xx km beyond the base distance. Determine f1(x)f^{-1}(x), writing your answer as an expression in xx.