# 6. Exponential and Logarithmic Equations

by M. Bourne

## Solving Exponential Equations using Logarithms

The logarithm laws that we met earlier are particularly useful for solving equations that involve exponents.

### Example 1

Solve the equation 3^{x} = 12.7.

Answer

We can estimate the answer before we start to be somewhere between `2` and `3`, because 3^{2} = 9 and 3^{3} = 27. But how do we find the answer?

First we take the logarithm of both sides of the given equation:

`log\ 3^x= log\ 12.7`

Now, using the 3rd log rule

log_{b}(x^{n}) =nlog_{b}x,

we have:

`x\ log\ 3 = log\ 12.7`

Now divide both sides by `log\ 3`:

`x= (log\ 12.7)/(log\ 3)``= 2.3135`

Is it correct? Checking in the original question, we have: `3^2.3135 = 12.7`. Checks okay. Also, our answer is between `2` and `3` as we estimated before.

### Example 2

Two populations of bacteria are growing at different rates. Their populations at time *t* are given by
`5^(t+2` and
*e*^{2t} respectively. At what time are the
populations the same?

Answer

This problem requires us to solve the equation:

5

^{t+2}=e^{2t}

We need
to use log_{e} because of the base *e* on the
right hand side.

ln (5

^{t+2}) = ln (e^{2t})

(

t+ 2) ln 5 = 2tlne

Now, ln *e* = 1, and we need to collect *t* terms
together:

tln 5 + 2 ln 5 = 2t

t(ln 5 − 2) = −2 ln 5

So

`t=(-2\ ln\ 5)/(ln\ 5-2)=8.241649476`

is the required time.

**Graph**

Graphs of `y=e^(2t)` (magenta) and `y=5^(t+2)` (green) showing the intersection point.

We can see on the graph that the 2 curves intersect at `t = 8.2`, as we found above.

### Exercises

(1) Solve `5^x= 0.3`

Answer

Taking log of both sides:

log 5

^{x}= log 0.3

So

`log\ 5^x=log\ 0.3`

`x\ log\ 5=log\ 0.3`

`x=(log\ 0.3)/(log\ 5)`

`=-0.748070363`

(2) Solve `3\ log(2x − 1) = 1`.

Answer

Divide both sides by 3:

`log(2x - 1) = 1/3`

So

`10^(1"/"3)=2x-1`

`x=1/2(10^(1"/"3)+1)`

`=1.577217345`

Always **check** your answers with your calculator!

(3) Solve for *x*:

`log_2 x + log_2 7 = log_2 21`

Answer

We first combine the 2 logs on the left into one logarithm.

`log_2\ 7x=log_2\ 21`

`7x=21`

`x=3`

To get the second line, we actually raise `2` to the power of the left side, and `2` to the power of the right side. We don't really "cancel out" the logs, but that is the effect (only if they have the same base).

(4) Solve for *x*:

`3\ ln\ 2+ln(x-1)=ln\ 24`

Answer

Recall that `3\ ln\ 2` means `3\ log_e\ 2`.

`3 ln\ 2+ln(x-1)=ln\ 24`

`ln\ 8+ln(x-1)=ln\ 24`

`ln\ 8(x-1)=ln\ 24`

We take "*e* to both sides":

`8(x-1)=24`

`x-1=3`

`x=4`

(5) [Reader's question.]

I have the following formula:

`S(n) = 5500\ log\ n + 15000` (Using base 10)

If I know *S*(*n*) = 40 million, How do I solve it?

Answer

40 000 000 = 5500 log* n* + 15000

log *n* = 7270.

This gives us simply *n* = 10^{7270} (using the log laws) which is pretty big!

(6) In the expression

`ln (x+2)^2 =\ 2,`

why is one of the answers not there when changed to

`2ln (x+2)=2,` thus `ln (x+2)=1,`

giving only one answer?

### Solution

One of the best ways to understand this problem is to see what's going on using some graphs.

Recall we can only find the logarithm of **positive** numbers.

Here's the graph of `y = ln (x+2)^2 - 2`, based on the first expression:

Graph of `y = ln (x+2)^2 - 2`.

We can see there are 2 roots (the 2 places where the graph cuts the `x`-axis.

There are two arms for the graph because we have squared the `(x+2)` term, meaning it will have value `> 0,` so we can take the `ln` of it with no problems (except at `x=-2,` of course, since it is undefined there).

Now let's look at the graph of `y = ln (x+2) - 1`, based on the final expression:

Graph of `y = ln (x+2) - 1`.

Now we only have one arm in the graph, and one place where the graph cuts the `x`-axis, thus giving us one solution for the equation `ln (x+2)=1,` which we find is (after taking `e` to both sides):

`x+2 = e`

`x = e - 2 = 0.718281828...`

So while the Log Law says we can write `ln (x+2)^2` as `2ln (x+2),` they are not really the same function.

## Application - World population growth

The population of the earth is growing at approximately `1.3%` per year. The population at the beginning of 2000 was just over `6` billion. After how many more years will the population double to `12` billion?

Answer

We need an expression for the population at time *t*.

After one year, the population will be `1.3%` higher than in 2000. (1.3% = 0.013)

Population after 1 year: `6\ "billion" × 1.013`.

Population after 2 years: `6\ "billion" × (1.013)^2`.

Population after 3 years: `6\ "billion" × (1.013)^3`.

So our population, *P*, after *t* years, is given by:

`P(t) = 6\ "billion" × (1.013)^t`

[In general, for any population growth,

`P(t) = P_0(1 + r)^t`

where *P*_{0} is the
population at time `t = 0`, *r* is the rate of growth per time period and *t* is the time.]

We are asked to find when the population doubles, so we need to solve:

`12\ 000\ 000\ 000 =` ` 6\ 000\ 000\ 000 × (1.013)^t`

This gives `2 = (1.013)^t`

Taking logarithms of both sides, we have:

`log\ 2 = log (1.013)^t`

Using the third log law, we have:

`log\ 2 = t\ log\ 1.013`

So

`t=(log\ 2)/(log\ 1.013)=53.66`

So it will take only about `54` years to double the world's population, if it continues to grow at the current rate.

When the world population is 12 billion, the net number of
people in the world will be increasing at the rate of
about 5 **per second**, if the growth rate is still 1.3%.
Currently, there are about 2.6 new people per second.
However, the rate of growth is expected to drop considerably
to about 0.5% within 50 years.

In 2001, the population of India passed **one billion**, making it the second country after China to reach that scary milestone.

### World population

Current world population is approximately:

**Loading...**

### Interactive applet - World Population

Go to the interactive World Population, which has comparisons between present, past and future population growth.

### Predicting world population

The following graph shows one of the estimates for world population growth during the 21st century. We see that the population will be 11 billion by about 2100! Think of our water quality, air pollution, global warming, social cohesion and lack of food. Surely this is one of the most important graphs in all of mathematics.

But I digress.

We are, of course, talking American English, here. The British billion has 12 zeroes (Well, even they have recently adopted the 9 zeroes billion...).

Graph of world population (billions), 1900 to 2100.

The world population is expected to exceed 11 billion by 2100. [Source]

This suggests a growth rate of about 0.6%, much lower than that experienced during the 20th century.

The equation for the above graph is

`P=6.1(1.006)^(t-2000)`, where

6.1 billion was the population in 2000;

the growth rate is represented by `1+6/100 = 1.006`; and

`t` is the time from the year 2000.

See a "live" world population estimation on the next page.