3.4 Logarithmic equations

Example 1

We solve the following equations for \displaystyle x:

1. \displaystyle 10^x = 537\quad has a solution \displaystyle x = \lg 537.
2. \displaystyle 10^{5x} = 537\quad gives \displaystyle 5x = \lg 537, i.e. \displaystyle x=\frac{1}{5} \lg 537.
3. \displaystyle \frac{3}{e^x} = 5 \quad . Multiplication of both sides with \displaystyle e^x and division by 5 gives \displaystyle \tfrac{3}{5}=e^x , which means that \displaystyle x=\ln\tfrac{3}{5}.
4. \displaystyle \lg x = 3 \quad. The definition gives directly \displaystyle x=10^3 = 1000.
5. \displaystyle \lg(2x-4) = 2 \quad. From the definition we have \displaystyle 2x-4 = 10^2 = 100 and it follows that \displaystyle x = 52.

Example 2

1. Solve the equation \displaystyle \,(\sqrt{10}\,)^x = 25.
   
   Since \displaystyle \sqrt{10} = 10^{1/2} the left-hand side is equal to \displaystyle (\sqrt{10}\,)^x = (10^{1/2})^x = 10^{x/2} and the equation becomes

   \displaystyle 10^{x/2} = 25\,\mbox{.}

   This equation has a solution \displaystyle \frac{x}{2} = \lg 25, ie. \displaystyle x = 2 \lg 25.
2. Solve the equation \displaystyle \,\frac{3 \ln 2x}{2} + 1 = \frac{1}{2}.
   
   Multiply both sides by 2 and then subtract 2 from both sides to get

   \displaystyle 3 \ln 2x = -1\,\mbox{.}

   Dividing both sides by 3 gives

   \displaystyle \ln 2x = -\frac{1}{3}\,\mbox{.}

   Now, the definition directly gives \displaystyle 2x = e^{-1/3}, so that

   \displaystyle x = {\textstyle\frac{1}{2}} e^{-1/3} = \frac{1}{2e^{1/3}}\,\mbox{.}

Example 3

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

   \displaystyle \lg 3^x = \lg 20\,\mbox{.}

   The left-hand side can be written as \displaystyle \lg 3^x = x \cdot \lg 3 giving

   \displaystyle x = \displaystyle \frac{\lg 20}{\lg 3} \quad ({}\approx 2\textrm{.}727)\,\mbox{.}

2. Solve the equation \displaystyle \ 5000 \cdot 1\textrm{.}05^x = 10\,000.
   
   Divide both sides by 5000 to get

   \displaystyle 1\textrm{.}05^x = \displaystyle \frac{ 10\,000}{5\,000} = 2\,\mbox{.}

   This equation can be solved by taking the lg logarithm of both sides of and rewriting the left-hand side as \displaystyle \lg 1\textrm{.}05^x = x\cdot\lg 1\textrm{.}05. Then

   \displaystyle x = \frac{\lg 2}{\lg 1\textrm{.}05} \quad ({}\approx 14\textrm{.}2)\,\mbox{.}

Example 4

1. Solve the equation \displaystyle \ 2^x \cdot 3^x = 5.
   
   The left-hand side can be rewritten using the laws of exponents giving \displaystyle 2^x\cdot 3^x=(2 \cdot 3)^x and the equation becomes

   \displaystyle 6^x = 5\,\mbox{.}

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

   \displaystyle x = \frac{\lg 5}{\lg 6}\quad ({}\approx 0\textrm{.}898)\,\mbox{.}

2. Solve the equation \displaystyle \ 5^{2x + 1} = 3^{5x}.
   
   Take logarithms of both sides and use the laws of logarithms to get

   \displaystyle \eqalign{(2x+1)\lg 5 &= 5x \cdot \lg 3\,\cr \Rightarrow 2x \cdot \lg 5 + \lg 5 &= 5x \cdot \lg 3\,\mbox{.}\cr}

   Collecting \displaystyle x to one side gives

   \displaystyle \eqalign{\lg 5 &= 5x \cdot \lg 3 -2x \cdot \lg 5\,\cr \Rightarrow \lg 5 &= x\,(5 \lg 3 -2 \lg 5)\,\mbox{.}\cr}

   The solution is then

   \displaystyle x = \frac{\lg 5}{5 \lg 3 -2 \lg 5}\,\mbox{.}

Example 5

Solve the equation \displaystyle \,\frac{6e^x}{3e^x+1}=\frac{5}{e^{-x}+2}.

Multiply both sides by \displaystyle 3e^x+1 and \displaystyle e^{-x}+2 to eliminate the denominators, so that

In this last step we have multiplied the equation by factors \displaystyle 3e^x+1 and \displaystyle 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 \displaystyle e^{-x} \cdot e^x = e^{-x + x} = e^0 = 1. If we treat \displaystyle e^x as the unknown variable, the equation is essentially a first order equation which has a solution

Taking logarithms then gives the answer:

Example 6

Solve the equation \displaystyle \,\frac{1}{\ln x} + \ln\frac{1}{x} = 1.

The term \displaystyle \ln\frac{1}{x} can be written as \displaystyle \ln\frac{1}{x} = \ln x^{-1} = -1 \cdot \ln x = - \ln x and then the equation becomes

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

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

   \textstyle (\ln x)^2 + \ln x -1
     &= \bigl( \ln x + \frac{1}{2} \bigr)^2 - \bigl(\frac{1}{2} \bigr)^2 - 1\\
     &= \bigl( \ln x + \frac{1}{2} \bigr)^2 - \frac{5}{4} \mbox{.}\\
 \end{align*}

Then by taking roots,

 \ln x = -\frac{1}{2} \pm \frac{\sqrt{5}}{2} \,\mbox{.}

This means that the equation has two solutions

 x= e^{(-1 + \sqrt{5})/2}
 \quad \mbox{or} \quad
 x= e^{-(1+\sqrt{5})/2}\,\mbox{.}

Example 7

Solve the equation \displaystyle \,\ln(4x^2 -2x) = \ln (1-2x).

For the equation to be satisfied the arguments \displaystyle 4x^2-2x and \displaystyle 1-2x must be be positive and equal i.e.

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

and taking the root. This gives

 \textstyle x= -\frac{1}{2}
 \quad\mbox{or}\quad
 x = \frac{1}{2} \; \mbox{.}

We now check if both sides of \displaystyle (*) are positive

• If \displaystyle x= -\tfrac{1}{2} then both are sides are equal to \displaystyle 4x^2 - 2x = 1-2x = 1-2 \cdot \bigl(-\tfrac{1}{2}\bigr) = 1+1 = 2 > 0.
• If \displaystyle x= \tfrac{1}{2} then both are sides are equal to \displaystyle 4x^2 - 2x = 1-2x = 1-2 \cdot \tfrac{1}{2} = 1-1 = 0 \not > 0.

So the logarithmic equation has only one solution \displaystyle x= -\frac{1}{2}.

Example 8

Solve the equation \displaystyle \,e^{2x} - e^{x} = \frac{1}{2}.

The first term can be written as \displaystyle e^{2x} = (e^x)^2. The whole equation is a quadratic with \displaystyle e^x as the unknown i.e.

The equation can be a little easier to manage if we write \displaystyle 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.