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}

   Collect \displaystyle x to one side

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

   The solution is

   \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

Note that since \displaystyle e^x and \displaystyle e^{-x} are always positive regardless of the value of \displaystyle x, in this latest 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

where we 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

where we can consider \displaystyle \ln x as a new unknown. We multiply both sides by \displaystyle \ln x (which is different from zero when \displaystyle x \neq 1) and this gives us a quadratic equation in \displaystyle \ln x

Completing the square on the left-hand side

   \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}\\
 \end{align*}

We continue by taking the root giving

 \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{och} \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 equal,

and also be positive. We solve the equation \displaystyle (*) by moving all of the terms to one side

and take the root. This gives that

 \textstyle x= -\frac{1}{2}
 \quad\mbox{and}\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

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

Complete the square for the left-hand side.

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

which gives solutions

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

Since \displaystyle \sqrt3 > 1 then \displaystyle \frac{1}{2}-\frac{1}{2}\sqrt3 <0 and 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 that

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

as the only solution to the equation.