Now that we’ve learned definition and properties of exponentiation, it is natural to start wondering how to solve equations containing exponents. They are called exponential equations.
While solving exponential equations, we”ll use the following fact:
$$a^x=a^y \Rightarrow x=y,$$
where $a$ is the base of the exponentiation and $x, y$ the exponents.
Let’s take a look at examples.
Example 1:
a) Solve $5^x=5^7$.
b) Solve $6^{1-x}=6^5$.
Solution:
a) The bases ($a=5$) are the same, so we immediately conclude that the exponents are also the same, i.e. $x=7$.
b) Here we also have the same bases ($a=6$), so the exponents also must be the same, i.e. $1-x=5$, from which it follows $x=6$.
Sometimes we will see equations with different bases. In that case, we need to convert one side (or both) of the equation, in order to use the mentioned rule.
Example 2:
a) Solve $4^x=16$.
b) Solve $3^{4x-1}=27$.
c) Solve $3^{x^2+3x}=81$.
Solution:
a) Noticing that $16=4^2$, we have $$4^x=4^2.$$ Now the bases are equal ($a=4$), so we conclude $x=2$.
b) Noticing that $27=3^3$, we have $$3^{4x-1}=3^3.$$ Therefore,
$$4x-1=3$$
$$4x=3+1$$
$$4x=4$$
$$x=1$$
We can always check if the number we’ve got is the solution.
Check:
$$3^{4 \cdot 1 – 1}= 3^{3}=27$$
Therefore, $x=1$ really is the solution.
c) Noticing that $81=3^4$, we have $$3^{x^2+3x}=3^4.$$
$$x^2+3x=4$$
$$x^2+3x-4=0$$
$$x_{1,2}=\frac{-3 \pm \sqrt{9+16}}{2}$$
$$x_{1,2}=\frac{-3 \pm 5}{2}$$
$$x_{1}=1, x_{2}=-4$$
Check:
For $x_{1}=1$: $3^{1^2+3 \cdot 1}=3^4=81$
For $x_{2}=-4$: $3^{(-4)^2+3 \cdot (-4)}=3^{16-12}=3^4=81$
Therefore, this equation has two solutions: $x_{1}=1, x_{2}=-4$.
In some equations we will need negative exponents:
Example 3: Solve $4^{5x+2}=\frac{1}{64}$.
Solution:
Noticing that $\frac{1}{64}=\frac{1}{4^3}=4^{-3}$, we have $$4^{5x+2}=4^{-3}$$
$$5x+2=-3$$
$$5x=-3-2$$
$$5x=-5$$
$$x=-\frac{5}{5}$$
$$x=-1$$
Check:
$$4^{5 \cdot (-1)+2}=4^{-3}=\frac{1}{64}$$
Therefore, $x=-1$ is the solution.
Example 4: Solve $4^{x}=-3$.
Solution:
Think about it; which exponent of positive number $4$ can give a negative number? For example, if we tried $x=0$, the result would be $4^0=1$. If we tried a negative number (for example $x=-3$), the result would be $4^{-3}=\frac{1}{64}$. If we tried any positive number $x$, the result would also be some positive number. Therefore, in all possible cases, the result of exponentiation is always a positive number! We can conclude that this equation has no solutions.
Example 5: Solve $2^x=31$.
Solution:
Here we have different bases and we can’t rearrange the equation to get the same bases, since $31$ is not the power of $2$. In this and similar examples, we”ll use a rule:
$$log_{a}x^{n}=nlog_{a}x.$$
Therefore,
$$log_{2}2^x=log_{2}31$$
$$x \cdot log_{2}2=log_{2}31$$
Using $log_{a}a=1$, we have
$$x=log_{2}31$$
Using the change of base rule, we have
$$x=\frac{ln31}{ln2} \approx 4.95.$$
Note: We could immediately use ln instead of $log_{2}$; we would get the same answer.
Example 6: Solve $5(2^{x-4})=203$.
Solution:
We will first divide the whole equation by $5$:
$$2^{x-4}=\frac{203}{5}$$
Since $\frac{203}{5}$ is not the power of $2$, we need to use a log rule:
$$ln(2^{x-4})=ln\left(\frac{203}{5}\right)$$
$$(x-4)ln2=ln\left(\frac{203}{5}\right)$$
$$x-4=\frac{ln\left(\frac{203}{5}\right)}{ln2}$$
$$x=\frac{ln\left(\frac{203}{5}\right)}{ln2}+4 \approx 9.34$$
Example 7: Solve $4^{6-9x}=\frac{1}{8^{x-3}}$.
Solution:
Noticing that $4=2^2$ and $8=2^3$, we have
$$(2^2)^{6-9x}=\frac{1}{(2^3)^{x-3}}$$
$$2^{2 \cdot (6-9x)}=\frac{1}{2^{3 \cdot (x-3)}}$$
We don’t want to have a fraction on the right side of the equation, so we apply the rule:
$$\frac{1}{a^n}=a^{-n}.$$
Using this gives
$$2^{2 \cdot (6-9x)}=2^{-3 \cdot (x-3)}$$
Since the bases are equal, we have
$$2 \cdot (6-9x)=-3 \cdot (x-3)$$
$$12-18x=-3x+9$$
$$-18x+3x=9-12$$
$$-15x=-3$$
$$x=\frac{-3}{-15}$$
$$x=\frac{1}{5}$$
Check:
Left side: $4^{6-9 \cdot \frac{1}{5}}=4^{6-\frac{9}{5}}=4^{\frac{21}{5}} \approx 337.79$
Right side: $\frac{1}{8^{\frac{1}{5}-3}}=\frac{1}{8^{-\frac{14}{5}}}=8^{\frac{14}{5}} \approx 337.79$
Therefore, $x=\frac{1}{5}$ is the solution.
Example 8: Solve $3e^x+6=486$.
Solution:
First we need to isolate the variable, i.e. $e^x$:
$$3e^x=486-6$$
$$3e^x=480$$
Dividing by $3$,
$$e^x=160$$
Since the bases aren’t equal, we will take the ln of both sides:
$$ln(e^x)=ln(160)$$
$$x \cdot ln(e)=ln(160)$$
$$x=ln(160) \approx 5.075$$
