Solving Exponential Equations

Suppose $2^m=2^n$. Could $m$ and $n$ be different? Look at the graph of $y=2^x$: it rises the whole way, so it never produces the same output twice. Every horizontal line meets it at most once, so each output comes from exactly one exponent. The same holds for any base $b>0$ with $b\ne 1$, so **if $b^m=b^n$, then $m=n$**. The base 1 is excluded for a reason: $1^3=1^7$, yet $3\ne 7$. The whole method rests on this rule, so the real skill is writing both sides as the same base to some power. Most equations use bases 2, 3 or 5 and their powers, and fractions and roots become powers too, using the negative and rational exponents from the earlier lessons.

When the two bases are both powers of the same number, rewrite both. In $8^{x+1}=4^{2x-3}$ neither 8 nor 4 is a power of the other, but both are powers of 2, so the equation becomes ${(2^3)}^{x+1}={(2^2)}^{2x-3}$, which is $2^{3x+3}=2^{4x-6}$. Matching exponents gives $3x+3=4x-6$, so $x=9$, as in the figure. Fractions are no obstacle: in ${\left(\frac{1}{9}\right)}^x=27^{x+2}$, write $\frac{1}{9}=3^{-2}$ and $27=3^3$ to get $3^{-2x}=3^{3x+6}$. Then $-2x=3x+6$, so $-5x=6$ and $x=-\frac{6}{5}$. If you are unsure which base to use, pick the smallest: 2 for 4, 8, 16, 32 and 64; 3 for 9, 27, 81 and 243; 5 for 25, 125 and 625. Sometimes more than one base works: $16^x=64$ can be $4^{2x}=4^3$ or $2^{4x}=2^6$, and both give $x=\frac{3}{2}$.

The matching rule needs a single power on each side, so $2^{x+2}-2^x=24$ cannot be matched as it stands, and the exponent laws do not combine a difference ($2^{x+2}-2^x$ is not $2^2$). The trick is the product law read backwards: $2^{x+2}=2^x\cdot 2^2=4\cdot 2^x$. Now both terms share the factor $2^x$, so factor it out: $2^x(4-1)=24$, so $3\cdot 2^x=24$, so $2^x=8$ and $x=3$. Check: $2^5-2^3=32-8=24$ ✓. Always factor out the lowest power so that the bracket holds whole numbers. In $5^{x+1}-5^{x-1}=120$ the lowest power is $5^{x-1}$, and $5^{x+1}=5^{x-1}\cdot 5^2$, so the equation becomes $5^{x-1}(25-1)=120$. Then $5^{x-1}=5$, so $x-1=1$ and $x=2$ (check: $125-5=120$ ✓).

Every power of a positive base is positive: $b^x>0$ for every real $x$. So once the power is isolated, look at the other side before doing anything else. $3^x=-9$ has no solution. It is tempting to write $x=-2$, but $3^{-2}=\frac{1}{9}$, which is positive. The same goes for $2^x+5=3$ (isolating gives $2^x=-2$) and for $7^x=0$. On a graph, the line $y=-9$ lies entirely below $y=3^x$, so they never meet. This check matters most in equations that factor like quadratics. In $4^x-2^x-12=0$, notice that $4^x={(2^2)}^x={(2^x)}^2$, so with $u=2^x$ the equation is $u^2-u-12=0$, which factors as $(u-4)(u+3)=0$. Then $2^x=4$ gives $x=2$, but $2^x=-3$ is impossible and must be rejected. The only solution is $x=2$ (check: $16-4-12=0$ ✓).

Try $2^x=10$. Ten is not a power of 2, so matching is impossible, and an exact answer needs logarithms, which come in Grade 12. You can still get as close as you like. First bracket it: $2^3=8$ and $2^4=16$, so $x$ is between 3 and 4, and closer to 3 because 10 is closer to 8. Then test decimals with a calculator, keeping track of which guesses land below 10 and which land above, as in the table. A graph gives the same answer: plot $y=2^x$ and $y=10$ on a graphing calculator or Desmos and read the $x$-coordinate of the intersection, $x\approx 3.32$. Graphing also handles equations with truly different bases, such as $2^x=3^{x-1}$, whose curves cross near $x\approx 2.71$.