Certain exponential equations arise that can be easily solved without resorting to logarithms.
The key to their solution lies in the recognition of various powers of integers. for example recognising that $81=3^4$81=34, or $125=5^3$125=53 or $256=2^8=4^4=16^2$256=28=44=162.
With a good understanding of the rules of exponents, in particular rules like $a^{-1}=\frac{1}{a}$a−1=1a and $a^{\frac{1}{2}}=\sqrt{a}$a12=√a, and more generally $a^{-m}=\frac{1}{a^m}$a−m=1am and $a^{\frac{1}{m}}=\sqrt[m]{a}$a1m=m√a, solutions can be readily found.
Here is a list of typical examples. Note that in each case it is possible to re-express certain numbers in the equation as powers. If this were not the case, then we would resort to a different strategy - one that involved the laws of logarithms.
Example 1
Solve $7^p=343$7p=343
| $7^p$7p | $=$= | $343$343 |
| $7^p$7p | $=$= | $7^3$73 |
| $\therefore$∴ $p$p | $=$= | $3$3 |
Example 2
Solve $2^{1-2x}=1024$21−2x=1024
| $2^{1-2x}$21−2x | $=$= | $1024$1024 |
| $2^{1-2x}$21−2x | $=$= | $2^10$210 |
| $1-2x$1−2x | $=$= | $10$10 |
| $2x$2x | $=$= | $-9$−9 |
| $\therefore$∴ $x$x | $=$= | $-4\frac{1}{2}$−412 |
Example 3
Solve $\left(3^2\right)^{x+1}=\frac{1}{81}$(32)x+1=181
| $\left(3^2\right)^{x+1}$(32)x+1 | $=$= | $\frac{1}{81}$181 |
| $\left(3^2\right)^{x+1}$(32)x+1 | $=$= | $\frac{1}{3^4}$134 |
| $3^{2x+2}$32x+2 | $=$= | $3^{-4}$3−4 |
| $2x+2$2x+2 | $=$= | $-4$−4 |
| $2x$2x | $=$= | $-6$−6 |
| $\therefore$∴ $x$x | $=$= | $-3$−3 |
Example 4
Solve $2^{2x}-20\left(2^x\right)+64=0$22x−20(2x)+64=0
| $2^{2x}-20\left(2^x\right)+64$22x−20(2x)+64 | $=$= | $0$0 |
| $\left(2^x\right)^2-20\left(2^x\right)+64$(2x)2−20(2x)+64 | $=$= | $0$0 |
| Setting $u$u | $=$= | $2^x$2x |
| $u^2-20u+64$u2−20u+64 | $=$= | $0$0 |
| $\left(u-16\right)\left(u-4\right)$(u−16)(u−4) | $=$= | $0$0 |
| $\therefore$∴ $u$u | $=$= | $4,16$4,16 |
Since $u=2^x$u=2x, we have $2^x=4$2x=4 and $2^x=16$2x=16, and thus $x=2$x=2 and $x=4$x=4.
Example 5
Solve $8^y=\sqrt[3]{32}$8y=3√32
| $8^y$8y | $=$= | $\sqrt[3]{32}$3√32 |
| $\left(2^3\right)^y$(23)y | $=$= | $\left(2^5\right)^{\frac{1}{3}}$(25)13 |
| $2^{3y}$23y | $=$= | $2^{\frac{5}{3}}$253 |
| $3y$3y | $=$= | $\frac{5}{3}$53 |
| $\therefore$∴ $y$y | $=$= | $\frac{5}{9}$59 |
Example 6
Solve $49\left(7^x\right)=2401\left(\sqrt{7}\right)^{2-4x}$49(7x)=2401(√7)2−4x
| $49\left(7^x\right)$49(7x) | $=$= | $2401\left(\sqrt{7}\right)^{2-4x}$2401(√7)2−4x |
| $7^2\times7^x$72×7x | $=$= | $7^4\left(7^{\frac{1}{2}}\right)^{2-4x}$74(712)2−4x |
| $7^{2+x}$72+x | $=$= | $7^4\times7^{1-2x}$74×71−2x |
| $7^{2+x}$72+x | $=$= | $7^{5-2x}$75−2x |
| $2+x$2+x | $=$= | $5-2x$5−2x |
| $\therefore$∴ $y$y | $=$= | $1$1 |
More Worked Examples (try these yourself and then check the answer)
Question 1
Solve the equation $\left(2^2\right)^{x+7}=2^3$(22)x+7=23 for $x$x.
Question 2
Consider the equation
$\left(2^x\right)^2-9\times2^x+8=0$(2x)2−9×2x+8=0
The equation can be reduced to a quadratic equation by using a certain substitution.
By filling in the gaps, determine the correct substitution that would reduce the equation to a quadratic.
Let $m=\left(\editable{}\right)^{\editable{}}$m=()
Solve the equation for $x$x by using the substitution $m=2^x$m=2x.
Question 3
Solve the equation $25^{x+1}=125^{3x-4}$25x+1=1253x−4.