We have seen that the value of a logarithm such as $\log_ba$logba is "the power to which $b$b must be raised in order to give $a$a".
That is, for a logarithmic equation of the form $\log_ba=x$logba=x, the value of $x$x is the power of $b$b that will give $a$a.
Algebraically, this means that the equation $\log_ba=x$logba=x is equivalent to the exponential equation $b^x=a$bx=a.
We can make use of this key relationship between logarithms and exponentials in order to solve logarithmic equations. First, rewrite the logarithmic equation in terms of exponentials, then use the properties of exponentials that we are familiar with to solve the equation.
Worked example
Solve the equation $\log_2x=7$log2x=7 for $x$x.
Think: This equation is in the form $\log_ba=c$logba=c, so we can start by rewriting it without logarithms.
Do: $\log_2x=7$log2x=7 is equivalent to $x=2^7$x=27, and so the solution is $x=128$x=128.
A logarithmic equation of the form $\log_ba=x$logba=x is equivalent to the exponential equation $b^x=a$bx=a.
We can use this to solve some equations by first rewriting them without logarithms.
Practice questions
Question 1
Solve $\log_45x=3$log45x=3 for $x$x.
Question 2
Solve $\log_{10}\left(3x+982\right)=3$log10(3x+982)=3 for $x$x.
Question 3
Solve $11\log_5\left(x-12\right)=33$11log5(x−12)=33 for $x$x.
Use the index laws and logarithm laws to see if you can simplify the expression first.
Practice question
Question 4
Solve $\log_{10}x-\log_{10}38=\log_{10}37$log10x−log1038=log1037 for $x$x.
Testing the solution for logarithmic equations
There are certain log equations where it is necessary to test the solution to determine if it really exists. Remember that the base and argument of a logarithm are always positive (and the base cannot be $1$1). Therefore we may find that, for some log equations, we may need to reject a solution because it results in a negative argument.
It may help to first determine the possible values the variable can take before solving the equation.
Practice question
Question 5
Solve the equation $\log_4\left(x+3\right)+\log_4\left(x-3\right)=2$log4(x+3)+log4(x−3)=2.