Solving equations with the logarithm

Equations with a logarithmic term

We first discuss equations where the logarithm appears in an equation, for example

4+3\cdot \log_5(4x)=10

The unknown ( $x$ ) appears in the logarithm. The strategy is to first isolate the logarithmic term:

\begin{array}{rll} 4+3\cdot \log_5(4x)&=&10\quad\vert -4,:3\\ \log_5(4x)&=&2\\ \end{array}

Now we have two choices:

Method 1 for solving logarithmic equations

Raise $5$ to the power of both sides, and simplify the resulting logarithmic term on the left:

\begin{array}{rll} \log_5(4x)&=&2 \quad\vert 5^{(.)}\\ 5^{\log_5(4x)} &=& 5^2\\ 4x&=&5^2 \end{array}

It is now simply enough to see that $x=\frac{25}{4}=\underline{6.25}$

Method 2 for solving logarithmic equations

From the equation

\log_5(4x)=2

we can argue directly that $4x=5^2$ , because only then do we have

\log_5(\underbrace{5^2}_{=4x})=2

So again we see that $x=\frac{25}{4}=\underline{6.25}$

Choose whatever method you prefer!!

Exercise 1

Solve the following equations:

$\log_5(x)=2$
$\log_4(2x)=-2$
$20\cdot \log_7(x^2)=\frac{80}{9}$
$1-\log_2(4x)+3=-7$
$\log_{10}(x^2-2x+1)=2$
$\log_3(x-1)=0$
$3\ln(2x)-1=5$

Solution

$x=25$
$x=\frac{1}{32}$
$x_1=1.541..., x_2=-1.541...$
$x=512$
$x_1=-9, x_2=11$
$x=2$
$x=3.69...$

Equations with an exponential term

We also require the logarithm for equations with an exponential term. For example,

3+2\cdot 5^x=75

The unknown appears in the exponent. To solve it, let us first isolate the exponential term:

\begin{array}{rll} 3+2\cdot 5^x&=&75\quad\vert -3,:2\\ 5^x &=& 36\\ \end{array}

Now again we have to methods how to go on.

Method 1 for solving exponential equations

Apply the logarithm base $5$ on both sides, and simplify the logarithmic term on the left side:

\begin{array}{rll} 5^x &=& 36 \quad\vert \log_5(.)\\ \log_5(5^x)&=&\log_5(36)\\ x &=&\log_5(36) \end{array}

To find the numerical value of this logarithm, we use

\log_5(36)=\frac{\log(36)}{\log(5)}=\underline{2.226...}

Method 2 for solving exponential equations

We use the fact that we can extract exponents inside a logarithm and multiply them with the logarithm. What we mean by this is that we can always write

\boxed{\log_b(c^n)=n\cdot \log_b(c)}

You can think of this theorem as a generalisation of the obvious fact that

\log_b(b^n)=n

Indeed, if we set $c=b$ we get

\log_b(b^n)=n\cdot \underbrace{\log_b(b)}_{=1}=n

We will prove this statement below, but let us now go on with solving the equation

5^x=36

by applying the logarithm base $10$ (or base $e$ ) on both sides of the equation, and then apply the theorem above to extract the $x$ from the inside of the logarithm:

\begin{array}{rll} 5^x &=& 36 \quad\vert \log(.)\\ \log(5^x)&=&\log(36)\\ x\cdot \log(5) &=&\log(36)\quad \vert :\log(5)\\ x &=&\frac{\log(36)}{\log(5)}\\ &=& \underline{2.226...} \end{array}

Again, you choose your preferred method.

Before we continue with solving some equations, let us quickly proof the theorem above as an exercise.

Exercise 2

Prove that $\log_b(c^n)=n\cdot \log_b(c)$ .

Solution

If the equation

\log_b(c^n)=n\cdot \log_b(c)

is correct, so must the equation

\underbrace{b^{\log_b(c^n)}}_{=c^n}=b^{n\cdot \log_b(c)}

The left-hand side is

b^{n\cdot \log_b(c)}=\left(b^{\log_b(c)}\right)^n=c^n

and we see that the equation is correct.

And now some equations to solve ...

Exercise 3

Solve the following equations
1. $3\cdot 1.5^x -1=17$
2. $4\cdot 2^{2x-1}+7=11.2$ .
3. $(3.2)^{\sqrt{x}}-3=5$
4. $4e^{x-1}=16$
5. $10^{\log_{10}(2x+1)}=4$
The world's population in 1950 was $2.5$ billion ( $2.5\cdot 10^9$ ). In 1970, it was $3.7$ billion. Assuming exponential growth, when will the population exceed $10$ billion for the first time? And when will this happen for a linear growth of the population?
Plutonium 239 is a man-made radioactive isotope used for nuclear explosives. It has a half-life of 24110 years, which means that every 24110 years half of the Plutonium in a sample decays, and half is left. How many years will it take until $1\%$ is left in the sample?

Solution

The solutions are
1. $x=4.419...$
2. $x=0.535...$
3. $x=3.196...$
4. $x=2.386...$
5. $x=1.5$
Exponential growth: Draw the diagram and find the function equation of the growth as a function of time $x$ :
$\begin{array}{rlll} \text{$y$}:& 2.5\cdot 10^9 &\xrightarrow[]{\cdot u} & 3.7\cdot 10^9 &\xrightarrow[]{\cdot u}& ... & \xrightarrow[]{\cdot u} & y\\ \text{$x$ (year)}:& 1950 & \xrightarrow[]{+20} & 1970 &\xrightarrow[]{+20}& ... &\xrightarrow[]{+20} & x\\ \end{array}$
The growth factor is
$u=\frac{3.7\cdot 10^9}{2.5\cdot 10^9}=1.48$
Thus, the function equation describing the growth is
$f(x)=2.5\cdot 10^9\cdot 1.48^{(x-1950)/20}$
Find $x$ such that
$2.5\cdot 10^9\cdot 1.48^{(x-1950)/20}=10\cdot 10^9$
We solve it the usual way
$\begin{array}{rll} 2.5\cdot 10^9\cdot 1.48^{(x-1950)/20}&=&10\cdot 10^9\quad\vert : (2.5\cdot 10^9)\\ 1.48^{(x-1950)/20} &=& 4 \quad\vert \log(.)\\ \frac{x-1950}{20}\cdot \log(1.48) &=& \log(4)\quad\vert :\log(1.48)\\ \frac{x-1950}{20} &=&\frac{\log(4)}{\log(1.48)}\quad\vert \cdot 20, +1950\\ x &=& \underline{2020.7\text{ years}} \end{array}$

linear growth:

\begin{array}{rlll} \text{$y$}:& 2.5\cdot 10^9 &\xrightarrow[]{+ u} & 3.7\cdot 10^9 &\xrightarrow[]{+ u}& ... & \xrightarrow[]{+ u} & y\\ \text{$x$ (year)}:& 1950 & \xrightarrow[]{+20} & 1970 &\xrightarrow[]{+20}& ... &\xrightarrow[]{+20} & x\\ \end{array}

The increment is

u=3.7\cdot 10^9-2.5\cdot 10^9=1.2\cdot 10^9

and thus

f(x)=2.5\cdot 10^9+1.2\cdot 10^9\cdot \frac{x-1950}{20}

Find $x$ with

2.5\cdot 10^9+1.2\cdot 10^9\cdot \frac{x-1950}{20}=10\cdot 10^9

So let's solve the equation

\begin{array}{rlll} 2.5\cdot 10^9+1.2\cdot 10^9\cdot \frac{x-1950}{20}&=&10\cdot 10^9\quad\vert -2.5\cdot 10^9, :(1.2\cdot 10^9)\\ \frac{x-1950}{20} &=& 6.25\quad\vert \cdot 20,+1950\\ x &=&\underline{2075\text{ years}} \end{array}

Let's say the initial mass of the Plutonium (at time $x=0$ ) is $m$ . If you want to be less abstract, just assume a value for $m$ , e.g. $m=100$ ( $m$ will eventually cancel anyway, so the result will be independent of $m$ ). Also, we set here $x=0$ for simplicity, but any value for the starting time will work as well, because we are interested in the duration.

\begin{array}{rlll} \text{$y$}:& m &\xrightarrow[]{\cdot 0.5} & 0.5m &\xrightarrow[]{\cdot 0.5}& ... & \xrightarrow[]{\cdot 0.5} & 0.01m\\ \text{$x$ (year)}:& 0 & \xrightarrow[]{+24110} & 24110 &\xrightarrow[]{+24110}& ... &\xrightarrow[]{+24110} & x\\ \end{array}

The growth factor is $u=0.5$ , so we have the following function equation describing the growth:

f(x)=m\cdot 0.5^{(x-0)/24410}

Find $x$ such that

m\cdot 0.5^{(x-0)/24410} = 0.01m

Notice now that we can cancel the $m$ , because it appears as a multiplication on both sides of the equation:

\begin{array}{rll} m\cdot 0.5^{x/24410} &=& 0.01m\quad\vert :m\\ 0.5^{x/24410}&=&0.01\quad\vert \log(.)\\ \frac{x}{24410}\cdot \log(0.5)&=&\log(0.01)\quad\vert \cdot 24110, :\log(0.5)\\ x &=& \underline{160183.37... \text{ years}} \end{array}

Note that we start with another time, say $x=1$ , the end result would be $160184.37... \text{ years}$ , and to find the duration, we would have to subtract $1$ from this result to get again the value above.