Solving Logarithmic Equations

Remember that logaM =x  means exactly the same thing as ax = M , that is, 

"logais the number to which you raise a in order to get M."

This is the key to solving equations in which logarithms appear.

For example:  Suppose we want to solve the equation  log2 y = 3.  Since this means exactly the same thing as  23   = y,  the equation pretty much solves itself!!  

Here's a slightly harder problem:   Solve the equation  log2 (5z+ 1) = 4.  Since this means exactly the same thing as  24   = 5z+1 , we need only solve the ordinary   equation 16 = 5z +1 .  So the solution is z = 3.

Exercises I

Solve  the following. 

  Solve log3 w= 4.  Solution: w=

 Solve  log3 (2x)=1. Solution: x=   

  Solve  log4 (x - 6)= 0. Solution: x =   

 Solve log10 (y - 2)= -1. Solution: y=    Hint:  What is 10 to the power -1?.

  log10 (2w+8)=3. Solution: w = 

You should see from these examples that the basic strategy is to convert the equation involving logarithms to one that doesn't involve logs (by using the equivalent exponential form of the equation), and then solving the converted equation.  In order to do this you will sometimes need to combine the logarithmic terms.


Example:  Solve  log2 (x + 1) +log2 (x) = 1.

The first step is to use properties of logarithms to combine the logarithmic terms. Using product rule we get:

log2 ((x + 1)x)  = 1    or    log2 (x2 + x)  = 1 

which is the same as

x2 + x = 21    or    x2 + x - 2= 0.

This is a quadratic equation, and you can easily solve it.  The solutions of this last equation are 

  x = 1   and   x = -2

BUT NOTE!!  ONLY  x = 1 is a solution of the original equation!  x = -2 cannot be a solution, because you can't take logs of negative numbers, so if you try to put x = -2 into the original logarithmic equation you would get 

log2 (-2 + 1) +log2 (-2) = 1

Neither logarithm makes sense, so -2 can't be a solution. 

This is the tricky part of solving logarithmic equations:  noticing and eliminating those pesky numbers that appear when you try to solve a given equation, which are not solutions. Remember, it is perfectly possible for a logarithmic equation not to have any solutions.

Exercises II

Determine the number of solutions for each of the following logarithmic equations. (Just the number of solutions, not their actual values.)  You might have to use the quadratic formula for some of these.  You will also have to be observant!

  The number of solutions of log2 (x + 1) +log2 (x - 1) = 3 is Hint:  Remember to combine the log terms and use what you know about simple quadratic equations. .

  The number of solutions of log2 (2 - x ) +log2 (2 + x) = 3 is = 

  The number of solutions of log2 (x2 + 1)  = 5 is  

Exercises III

Each of the following has exactly one solution.  Find it.  Remember, your converted equation may have a solution that is not a solution of the original equation.   

log2 (x - 2) +log2 (x - 3) = 1  has solution x =

 log10 (x + 3) +log10 (x) = 1  has solution x =

Precalculus Tutorials, B. Kaskosz and L. Pakula, 2002.