2 Solving equations involving logarithms and exponentials

To solve equations which involve logarithms or exponentials we need to be aware of the basic laws which govern both of these mathematical concepts. We illustrate by considering some examples.

Example 6

Solve for the variable $x$ :

1. $3=10^x$ ,
2. $10^{x∕4}=log3$ ,
3. $\frac{1}{17-{\text{e}}^x}=4$

Solution

1. Here we take logs (to base 10 because of the term $10^x$ ) of both sides to get

   $log3=log10^x=xlog10=x$

   where we have used the general property that ${log}_a{A}^k=k{log}_{a}A$ and the specific property that $log10=1$ . Hence $x=log3$ or, in numerical form, $x=0.47712$ to 5 d.p.

2. The approach used in (1) is used here. Take logs of both sides: $log\left(10^{x∕4}\right)=log\left(log3\right)$

   $\text{that is}\frac{x}{4}log10=log\left(log3\right)=log\left(0.4771212\right)=-0.3213712$

   So, since $log10=1$ , we have $x=4\left(-0.3213712\right)=-1.28549\text{to 5 d.p.}$

3. Here we simplify the expression before taking logs.

   $\frac{1}{17-{\text{e}}^x}=4\text{implies}1=4\left(17-{\text{e}}^x\right)$

   or $4{\text{e}}^x=4\left(17\right)-1=67$ so ${\text{e}}^x=16.75$ . Now taking natural logs of both sides (because of the presence of the ${\text{e}}^x$ term) we have:

   $ln\left({\text{e}}^x\right)=ln\left(16.75\right)=2.8183983$

   But $ln\left({\text{e}}^x\right)=xln\text{e}=x$ and so the solution to $\frac{1}{17-{\text{e}}^x}=4$ is $x=2.81840$ to 5 d.p.

Task!

Solve the equation ${\left({\text{e}}^x\right)}^2=50$

First solve for ${\text{e}}^x$ by taking square roots of both sides:

Answer

Now take logarithms to an appropriate base to find $x$ :

Answer

Task!

Solve the equation  ${\text{e}}^{2x}=17{\text{e}}^x$

First simplify the expression as much as possible (divide both sides by ${\text{e}}^x$ ):

Answer

Now complete the solution for $x$ :

Answer

Example 7

Find $x$ if $10^x-5+6\left(10^{-x}\right)=0$

Solution

We first simplify this expression by multiplying through by $10^x$ (to eliminate the term $10^{-x}$ ):

$10^x\left(10^x\right)-10^x\left(5\right)+10^x\left(6\left(10^{-x}\right)\right)=0$

or

${\left(10^x\right)}^2-5\left(10^x\right)+6=0\text{since}10^x\left(10^{-x}\right)=10^0=1$

We realise that this expression is a quadratic equation . Let us put $y=10^x$ to give

$y^2-5y+6=0$

Now, we can factorise to give

$\left(y-3\right)\left(y-2\right)=0\text{so that}y=3\text{or}y=2$

For each of these values of $y$ we obtain a separate value for $x$ since $y=10^x$ .

Case 1 If $y=3$ then $3=10^x$ implying $x=log3=0.4771212$

Case 2 If $y=2$ then $2=10^x$ implying $x=log2=0.3010300$

We conclude that the equation $10^x-5+6\left(10^{-x}\right)=0$ has two possible solutions for $x$ : either $x=0.4771212$ or $x=0.3010300$ , to 7 d.p.

Task!

Solve $2{\text{e}}^{2x}-7{\text{e}}^x+3=0$ .

First write this equation as a quadratic in the variable $y={\text{e}}^x$ remembering that ${\text{e}}^{2x}\equiv {\left({\text{e}}^x\right)}^2$ :

Answer

Now solve the quadratic for $y$ :

Answer

Finally, for each of your values of $y$ , find $x$ :

Answer

The temperature $T$ , in degrees C, of a chemical reaction is given by the formula

$T=80{\text{e}}^{0.03t}\times t\ge 0,$  where $t$ is the time, in seconds.

Calculate the time taken for the temperature to reach $150^{\circ }$ C .

Answer