Solving exponential and logarithmic equations

Taking logs to solve a^x = b, equations with e^x and ln, hidden quadratics in e^x or a^x, and log equations solved with the laws — always checking which roots are valid.

Pure · Exponentials & logarithms

Solving exponential and logarithmic equations

Taking logs to solve a^x = b, equations with e^x and ln, hidden quadratics in e^x or a^x, and log equations solved with the laws — always checking which roots are valid.

Why it works

When the unknown is stuck in the exponent, take logs of both sides and use the power law to bring it down. To solve

a^x = b

a^x = b \;\Rightarrow\; \log a^x = \log b \;\Rightarrow\; x\log a = \log b \;\Rightarrow\; x = \frac{\log b}{\log a}.

Any base works; for

e^x = b

the natural choice is

\ln

, giving

x = \ln b

directly.

Hidden quadratics. Equations like

e^{2x} - 5e^{x} + 6 = 0

2^{2x} - 9\cdot 2^{x} + 8 = 0

look exponential but become quadratics under a substitution: let

u = e^{x}

(so

e^{2x} = u^2

) or

u = 2^{x}

. Solve for

u

, then go back with logs — and discard any non-positive $u$ , because

e^x

and

2^x

are always positive.

Log equations. When the unknown is inside logs, use the laws to collapse each side to a single log, then either equate the arguments (if the bases and the "

\log =\log

" form match) or rewrite in index form:

\log_a f(x) = c \;\Rightarrow\; f(x) = a^{c}.

Always check your roots. A log is only defined for a positive argument, so any candidate that makes any log in the original equation take the log of zero or a negative number must be rejected — it is not a genuine solution even though it solves the quadratic.