Solving Exponential Equations

To solve exponential equations, we need to consider the rule of exponents. These rules help us a lot in solving these type of equations.

In solving exponential equations, the following theorem is often useful:

If $a$ is a non-zero constant and $a^x = a^y,$ then $x = y.\ _\square$

We have

$$\begin{aligned} a^x&=a^y, a\neq 1 \\ \frac{a^x}{a^y} &= 1 \\ a^{x-y} &= 1 \\ x-y &= 0\\ x &= y.\ _\square \end{aligned}$$

Here is how to solve exponential equations:

1. Manage the equation using the rule of exponents and some handy theorems in algebra.
2. Use the theorem above that we just proved.

Solve $\displaystyle{ \frac{1}{5^{x-1}} = 125} .$

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Making the bases on both sides equal to 5 gives

$$\begin{aligned} \frac{1}{5^{x-1}} &= 125 \\ 5^{-(x-1)}&=5^3 \\ -(x-1) &= 3 \\ x &= -2. \ _\square \end{aligned}$$

Solve $4^{x-3} = 0.125 .$

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Converting the bases of both sides to 2 gives

$$\begin{aligned} 4^{x-3} &= 0.125 \\ 4^{x-3} &= \frac{125}{1000} \\ 2^{2x-6} &= \frac{1}{8} \\ 2^{2x-6} &= 2^{-3} \\ 2x-6&= -3 \\ x&= \frac{3}{2}. \ _\square \end{aligned}$$

Find the value of $x$ when $4^x = 16.$

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We have $4^x = 16 \Rightarrow 4^x = 4^2.$ Then the theorem "if $a$ is a non-zero constant and $a^x = a^y,$ then $x = y$" gives $x = 2.\ _\square$

If $8^x = 2$, what is $x$?

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We have

\[\begin{align}
8^x & = 2\\ \big(2^3\big)^{x} & = 2\\ 2^{3x} & = 2^{1}. \end{align}\]

Equating the powers, we get

\[\begin{align}
3x & = 1\\ x & = \dfrac{1}{3}.\ _\square \end{align}\]

If $6^x - 1=0,$ what is $x?$

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We have

$$\begin{aligned} 6^x-1&=0 \\ 6^x &= 1\\ 6^x &= 6^0\\ x &= 0.\ _\square \end{aligned}$$

If $(8)(9^x) = 9^x,$ what is $x?$

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We have

$$\begin{aligned} (8)(9^x) &= 9^x \\ (8)(9^x) - 9^x &= 0 \\ (7)9^x &= 0 \\ 9^x &= 0. \end{aligned}$$

Then the theorem "if $a \neq 0$ and $a^x=0,$ then $x = \phi$" gives $x = \phi.\ _\square$

If the bases are different, there are still techniques for solving these exponential equations. If the bases are powers of a common base, we need only convert one or both bases to the common base and proceed using the "Same Base" case.

Solve $4^{3x} =8^{x-1}.$

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We see that while 4 and 8 are different bases, they are both powers of a common base, namely 2. We'll proceed by rewriting 4 and 8 in terms of their common base:

$$\begin{aligned} 4^{3x} &= 8^{x-1} \\ \big(2^2\big)^{3x} &=\big (2^3\big)^{x-1} \\ 2^{6x} &= 2^{3x-3} \\ 6x&=3x-3 \\ x&= -1. \ _\square \end{aligned}$$

Solve for $x:$ $\dfrac{8^{4x - \sqrt{x}}}{{16}^{2x + \sqrt{x}}} = 2^{2\sqrt{x}}$

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We have

\[\begin{align}
\dfrac{8^{4x - \sqrt{x}}}{{16}^{2x + \sqrt{x}}} & = 2^{2\sqrt{x}}\\ \\ \dfrac{{\big(2^3\big)}^{4x - \sqrt{x}}}{{{\big(2^4\big)}}^{2x + \sqrt{x}}} & = 2^{2\sqrt{x}}\\ \\ \dfrac{2^{12x - 3\sqrt{x}}}{2^{8x + 4\sqrt{x}}} & = 2^{2\sqrt{x}}\\ \\ 2^{12x - 3\sqrt{x} - 8x - 4\sqrt{x}} & = 2^{2\sqrt{x}}\\ \\ 2^{4x - 7\sqrt{x}} & = 2^{2\sqrt{x}}. \end{align}\]

Equating the powers gives

\[\begin{align}
4x - 7\sqrt{x} & = 2\sqrt{x}\\ 4x & = 9\sqrt{x}\\ \sqrt{x} & = \dfrac{9}{4}\\ x & = \dfrac{81}{16}.\ _\square \end{align}\]

Solve $$\frac{27^{3x-2}}{243} = 81^{3x-6} .$$

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Reducing the bases of 27, 81, and 243 on both sides to 3 yields

$$\begin{aligned} \frac{27^{3x-2}}{243} &= 81^{3x-6} \\ \frac{3^{9x-6}}{3^5} &= 3^{12x-24} \\ 3^{9x-6-5} &= 3^{12x-24} \\ 3^{9x-11} &= 3 ^{12x-24} \\ 9x-11 &= 12x-24 \\ 3x &= 13 \\ x &= \frac{13}{3}. \ _\square \end{aligned}$$

Unfortunately, it won't always be possible to convert to a common base as we did in the examples above.

For instance, in solving $5^{x} = 3^{x + 2}$, we note that 5 and 3 are not powers of a nice common base. In this case, we'll need to make use of logarithms.

Solve $5^{x} = 3^{x + 2}.$

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We have

$$\begin{aligned} 5^{x} &= 3^{x + 2} \\ \log_{10}(5^{x}) &= \log_{10}\big(3^{x + 2}\big) \\ x\log_{10}5 &= (x + 2)\log_{10}3 \\ x\log_{10}5 &= x\log_{10}3 + 2\log_{10}3 \\ x\log_{10}5 - x\log_{10}3 &= 2\log_{10}3 \\ x\big(\log_{10}5 - \log_{10}3\big) &= 2\log_{10}3. \end{aligned}$$

Therefore,

$$x = \frac{2\log_{10}3}{\log_{10}5 - \log_{10}3} \approx 4.3013 .\ _\square$$

Given $1728 = 2^a.3^b$, find positive integers $a$ and $b$.

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We have

\[\begin{align}
1728 & = {12}^{3}\\ & = {(4 × 3)}^{3}\\ & = 4^3 × 3^3\\ & = {(2^2)}^3 × 3^3\\ & = 2^6 × 3^3\\ \Rightarrow a & = 6, \ b = 3.\ _\square \end{align}\]

Solve $3^{x^2} = 3^{x}.$

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Since both sides of the equation have the same base, their exponents must also be the same:

$$\begin{aligned} 3^{x^2} &= 3^{x} \\ x^2 &=x \\ x^2-x&=0\\ x(x-1)&=0\\ \Rightarrow x&=0,1. \ _\square \end{aligned}$$

If $2^x \cdot 3^y \cdot 5^z = 45,$ what is value of $x+y+z?$

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Observe that $45$ can be factorized as follows:

$$45 = 3^2\cdot5=2^0\cdot3^2\cdot5^1.$$

Then $x=0, y=2, z=1,$ which gives $x+y+z=0+2+1=3.$ $_\square$

If $x^x \cdot y^y = 108 ,$ what is value of $x+y ?$

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$108$ can be factorized as follows:

$$108 = 2^2 \cdot 3^3 .$$

This implies that either $x=2, y=3$ or $x=3, y=2.$ Therefore the answer is $x+y=2+3=5. \ _ \square$

If $\frac{2^5}{2^3} \cdot 3^0 \cdot 3^1 \cdot 3^2 = 2^x \cdot 3^y ,$ what is value of $x+y?$

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$\frac{2^5}{2^3}$ can be rewritten as

$$\frac{2^5}{2^3} = {2}^{5-3}=2^2. \qquad (1)$$

$3^0 \cdot 3^1 \cdot 3^2$ can be rewritten as

$$3^0 \cdot 3^1 \cdot 3^2 = 3^{0+1+2}=3^3. \qquad (2)$$

From $(1)$ and $(2),$ $x$ and $y$ are

$$\begin{aligned} \frac{2^5}{2^3} \cdot 3^0 \cdot 3^1 \cdot 3^2 &= 2^2 \cdot 3^3 \\ &= 2^x \cdot 3^y \\ \Rightarrow x&= 2,\ y=3. \end{aligned}$$

Hence, $x+y= 2+3 = 5. \ _ \square$

An exponential equation is one in which a variable occurs in the exponent. If both sides of the equation have the same base, then the exponents on both sides are also the same:

$$a^x=a^y \implies x=y .$$

Here is a list of some rules concerning exponential functions:

$\quad (1)~~a^m \times a^n = a^{m+n}$
$\quad (2)~~a^m \div a^n = a^{m-n}$
$\quad (3)~~(a^m)^n = a^{mn}$
$\quad (4)~~(ab)^n=a^{n}b^{n}$
$\quad (5)~~a^0=b^0$
$\quad (6)~~1^m=1^n,$

where $a\neq0$ and $b\neq0.$ Always be cautious of $(5)$ and $(6)$; never forget to check if plugging a zero in an exponent works, or if there are any bases that are equal to 1.