Solving Exponential Equations

Use the one-to-one property of a valid base

For $b>0$ and $b\ne1$,

This follows because a real exponential with that base is strictly monotone: increasing for $b>1$, decreasing for $0<b<1$.

Solve

Rewrite using base 3:

Equate exponents to obtain $2x-2=6x+3$, so $x=-5/4$. The exponents of 3 are both $-9/2$ at that value. This verifies the result without approximating the powers.

Base one is an exception: $1^u=1^v$ says nothing about whether $u=v$. Negative bases do not define ordinary real exponentials at every real exponent, so they need separate domain treatment rather than this rule.

Isolate a positive exponential before taking logarithms

Solve

Divide by 5, then take a logarithm of both positive sides:

Hence

This exact expression is preferable to rounding intermediate logarithms. Substitution works because it makes $3x-1=\log_2(17/5)$, so the power equals $17/5$ exactly.

If isolation instead produced $2^{3x-1}=-17/5$, there would be no real solution. Every real power of a positive base is positive. Do not apply a real logarithm to a nonpositive value merely because the equation has an unknown to solve.

The OpenStax exponential-equation treatment covers the one-to-one and logarithm methods used here.

Convert a repeated exponential into a quadratic

Solve

Since $4^x=(2^x)^2$, let $u=2^x$. Crucially, $u>0$. The equation becomes

Both candidates are positive. Return to the original variable:

Thus the complete real solution set is $\{0,2\}$.

For the related equation $4^x+3\cdot2^x-4=0$, substitution gives $(u+4)(u-1)=0$. The candidate $u=-4$ is impossible because $2^x>0$. Only $u=1$ remains, giving $x=0$.

The quadratic formula can generate algebraic candidates for the substituted variable, but it cannot override that variable’s range. Record that range as soon as the substitution is made.

Handle reciprocals through the same positive variable

Consider

With $u=2^x>0$, this becomes $u+1/u=5/2$. Multiply by $2u$, which is positive and nonzero:

The positive possibilities are $u=1/2$ and $u=2$, so $x=-1$ and $x=1$. Substitution returns $1/2+2=5/2$ in either order.

The expression $u+1/u$ is at least 2 for positive $u$, since $(u-1)^2\ge0$. Therefore an equation $2^x+2^{-x}=1$ has no real solution before any quadratic calculation is attempted. Range reasoning can shorten the work and explain why a discriminant test gives no acceptable result.

Reverse an inequality for a decreasing base

Solve

Write $9=(1/3)^{-2}$. Because the base $1/3$ gives a decreasing function, the order of the exponents is reversed:

Equivalently, taking natural logarithms and dividing by $\ln(1/3)<0$ reverses the inequality. Both methods encode the same monotonicity.

For a base greater than one, exponent inequalities keep their direction. If a further step divides by a negative coefficient of $x$, that creates another reversal. Track each reason separately rather than memorising the appearance of a final answer.

Find a numerical solution with a completeness argument

The equation

does not allow the logarithm to be distributed over its left-side sum. Define $F(x)=2^x+3^x-10$. Both exponential terms are continuous and strictly increasing, so $F$ is continuous and strictly increasing on the reals.

At 1, $F(1)=-5$; at 2, $F(2)=3$. There is a root between them, and strict increase makes it unique. Bisection then repeatedly halves an interval whose endpoint values have opposite signs.

The endpoint values are $2^{1.72}+3^{1.72}\approx9.911172$ and $2^{1.73}+3^{1.73}\approx10.007180$, so they lie below and above 10 respectively. Its midpoint 1.725 has an absolute error below 0.005. More bisection steps produce a smaller bound.

A calculator’s decimal answer becomes much more informative when accompanied by the defining equation, an interval and a uniqueness argument. Without those, a numerical method might have found one root while missing others.

Distinguish real time from an integer index

An equation solved over the reals may describe a crossing between observations. Suppose an exponential quantity is measured only at completed periods and a real calculation gives a threshold index $n\ge4.3$. The first eligible integer is 5, not 4.

That conclusion requires the inequality direction and monotonicity to be correct. Check the values at 4 and 5 in the original model. For exact term membership, rather than a threshold, a noninteger solution means no indexed term equals the target. Geometric Sequences develops this distinction in detail.

Related topics

Use Solving Logarithmic Equations for equations in which the variable appears inside a logarithm. Exponential Growth and Decay applies these tools to fitted and competing models.