Exponential and logarithmic equations and inequalities

Exponential equations: three methods

Match the bases. If both sides can be written as powers of the same base, set the exponents equal, because $a^u = a^v$ forces $u = v$ when $a > 0$ and $a \neq 1$. Typical bases are 2 (for 4, 8, 16, 32), 3 (for 9, 27, 81) and 5 (for 25, 125). Fractions and roots fit too: $\frac{1}{4} = 2^{-2}$ and $\sqrt{8} = 2^{\frac{3}{2}}$.

Substitute. When the equation mixes $a^{2x}$ and $a^x$, or $a^{x+1}$ and $a^x$, set $t = a^x$. Then $a^{2x} = t^2$ and $a^{x+1} = a \cdot t$, and the equation becomes a polynomial in $t$. Solve it, then return to $x$. Because $a^x$ is always positive, only positive values of $t$ give solutions; a zero or negative root in $t$ is discarded.

Take logarithms. When the bases cannot be matched, as in $3^x = 7$, the answer is a logarithm: $x = \log_3 7$. On a test without a calculator, options are then written as logarithms, or you need only to locate the value: $3^1 < 7 < 3^2$ puts $x$ between 1 and 2.

Logarithmic equations: domain first

A logarithmic equation has an existence condition: every argument must be positive. Write it down before solving. Then:

1. Use the rules to combine each side into a single logarithm with one base.
2. Convert: $\log_a E = c$ becomes $E = a^c$; $\log_a E = \log_a F$ becomes $E = F$.
3. Solve the resulting algebraic equation.
4. Keep only the roots that satisfy the existence condition.

Step 4 is not optional. Combining $\log_a x + \log_a(x - 2)$ into $\log_a\big(x(x - 2)\big)$ quietly enlarges the domain, since $x(x - 2) > 0$ also holds for negative $x$, where the original logarithms do not exist. Roots from that enlarged domain are extraneous.

Inequalities: watch the base

Both $a^x$ and $\log_a x$ are increasing when $a > 1$ and decreasing when $0 < a < 1$. So after rewriting both sides with the same base:

Base	$a^u < a^v$ means	$\log_a u < \log_a v$ means
$a > 1$	$u < v$	$u < v$ (with $u > 0$)
$0 < a < 1$	$u > v$	$u > v$ (with $v > 0$)

With a base between 0 and 1, the inequality sign flips when you drop the base. For a logarithmic inequality, intersect the result with the existence condition; for an exponential one there is no domain restriction, since $a^x$ exists for every $x$.

Some inequalities are settled by positivity alone. $2^x > -3$ holds for every real $x$, and $2^x < 0$ for none, because an exponential with positive base is always positive.

Checking costs seconds

Substitute each answer into the original equation, not into a rewritten one. For exponential equations, choose a root that makes the arithmetic easy; for logarithmic ones, check first that every argument is positive, then that the two sides agree. For an inequality, test one value inside your solution set and one outside.

Reading the options

When the options mix a correct root with a set that includes an extraneous one, the extraneous root is usually negative or makes some argument zero. When the options for an inequality differ only in the direction of the sign, the question is testing whether you noticed a base less than 1.