3.4 Logarithmic equations

Example 1

We solve the following equations for x:

1. 10x=537 has a solution x=lg537.
2. 105x=537 gives 5x=lg537, i.e. x=51lg537.
3. 3ex=5. Multiplication of both sides with ex and division by 5 gives 53=ex, which means that x=ln53.
4. lgx=3. The definition gives directly x=103=1000.
5. lg(2x−4)=2. From the definition we have 2x−4=102=100 and it follows that x=52.

Example 2

1. Solve the equation (10)x=25 .
   
   Since 10=1012  the left-hand side is equal to (10)x=(1012)x=10x2  and the equation becomes

   10x2=25.

   This equation has a solution x2=lg25, ie. x=2lg25.
2. Solve the equation 23ln2x+1=21.
   
   Multiply both sides by 2 and then subtract 2 from both sides to get

   3ln2x=−1.

   Dividing both sides by 3 gives

   ln2x=−31.

   Now, the definition directly gives 2x=e−13 , so that

   x=21e−13=12e13.

Example 3

1. Solve the equation 3x=20.
   
   Take logarithms of both sides to get

   lg3x=lg20.

   The left-hand side can be written as lg3x=xlg3 giving

   x=lg3lg20(2.727).

2. Solve the equation  50001.05x=10000.
   
   Divide both sides by 5000 to get

   1.05x=500010000=2.

   This equation can be solved by taking the lg logarithm of both sides of and rewriting the left-hand side as lg1.05x=xlg1.05. Then

   x=lg2lg1.05(14.2).

Example 4

1. Solve the equation  2x3x=5.
   
   The left-hand side can be rewritten using the laws of exponents giving 2x3x=(23)x and the equation becomes

   6x=5.

   This equation is solved in the usual way by taking logarithms giving

   x=lg6lg5(0.898).

2. Solve the equation  52x+1=35x.
   
   Take logarithms of both sides and use the laws of logarithms to get

   (2x+1)lg52xlg5+lg5=5xlg3=5xlg3.

   Collecting x to one side gives

   lg5lg5=5xlg3−2xlg5=x(5lg3−2lg5).

   The solution is then

   x=lg55lg3−2lg5.

Example 5

Solve the equation 6ex3ex+1=5e−x+2.

Multiply both sides by 3ex+1 and e−x+2 to eliminate the denominators, so that

In this last step we have multiplied the equation by factors 3ex+1 and e−x+2. Both of these factors are different from zero, so this step cannot introduce new (spurious) roots of the equation.

Simplify both sides of the equation to get

Here we have used e−xex=e−x+x=e0=1. If we treat ex as the unknown variable, the equation is essentially a first order equation which has a solution

Taking logarithms then gives the answer:

x=ln31=ln3−1=−1ln3=−ln3.

Example 6

Solve the equation 1lnx+lnx1=1.

The term lnx1 can be written as lnx1=lnx−1=−1lnx=−lnx and then the equation becomes

We multiply both sides by lnx (which is different from zero when x=1, and x=1 is clearly not a solution) and this gives us a quadratic equation in lnx:

(lnx)2+lnx−1=0.

Completing the square on the left-hand side we see that

(lnx)2+lnx−1=lnx+212−212−1=lnx+212−45.

Then by taking roots,

lnx=−2125.

This means that the equation has two solutions

x=e(−1+5)2orx=e−(1+5)2.

Example 7

Solve the equation ln(4x2−2x)=ln(1−2x).

For the equation to be satisfied the arguments 4x2−2x and 1−2x must be be positive and equal i.e.

4x2−2x=1−2x.

()

We solve the equation () by moving all of the terms to one side

and taking the root. This gives

We now check if both sides of () are positive

• If x=−21 then both are sides are equal to 4x2−2x=1−2x=1−2−21=1+1=20 .
• If x=21 then both are sides are equal to 4x2−2x=1−2x=1−221=1−1=00.

So the logarithmic equation has only one solution x=−21.

Example 8

Solve the equation e2x−ex=21.

The first term can be written as e2x=(ex)2. The whole equation is a quadratic with ex as the unknown i.e.

The equation can be a little easier to manage if we write t instead of \displaystyle e^x, so that we try and solve

Completing the square on the left-hand side gives

   \textstyle \bigl(t-\frac{1}{2}\bigr)^2 - \bigl(\frac{1}{2}\bigr)^2
     &= \frac{1}{2}\,\\
   \Rightarrow \bigl(t-\frac{1}{2}\bigr)^2
     &= \frac{3}{4}\,\mbox{.}\\
 \end{align*}

so that

 t=\frac{1}{2} - \frac{\sqrt{3}}{2}
 \quad\mbox{or}\quad
 t=\frac{1}{2} + \frac{\sqrt{3}}{2} \, \mbox{.}

Since \displaystyle \sqrt3 > 1, \displaystyle \frac{1}{2}-\frac{1}{2}\sqrt3 <0. Therefore it is only \displaystyle t= \frac{1}{2}+\frac{1}{2}\sqrt3 that provides a solution to the original equation because \displaystyle e^x is always positive. Taking logarithms finally gives

 x = \ln \Bigl(\,\frac{1}{2}+\frac{\sqrt3}{2}\,\Bigr)

as the only solution to the equation.