Change of base formula for logarithms
Change of base formula for logarithms
It’s easier for us to evaluate logs of base ???10??? or base ???e???, because calculators usually have ???\log??? and ???\ln??? buttons for these.
When the base is anything other than ???10??? or ???e???, we can use the change of base formula.
???\log_ab=\frac{\log_cb}{\log_ca}???
Notice that, given a log with a base of ???a??? and an argument of ???b???, we can pick any value that we’d like to be the new base, ???c???. Which is really helpful, because we can pick a new base of ???10??? or ???e???, if either one is convenient for us.
How to use the change of base formula to simplify expressions with logs (logarithms)
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Converting a log with a different base to a log with base 10
Example
Estimate the log to four decimal places.
???\log_54???
We can use the change of base formula.
???\log_ab=\frac{\log_cb}{\log_ca}???
???\log_54=\frac{\log_{10}4}{\log_{10}5}???
Now we can use a calculator to evaluate the answer.
???\log_54\approx\frac{0.6021}{0.6990}???
???\log_54 \approx 0.8614???
Realize that we can also work backwards, backing our way into the change of base formula.
Example
Simplify the expression into a single real number without using a calculator.
???\frac{\log{625}}{\log{25}}???
If we use the change of base formula, we can rewrite this expression in the form ???\log_ab???.
???\log_{25}{625}???
Then, using the general form for logarithms, we can say that the value of this log is given by ???x???, and therefore
???25^x=625???
Now we want to rewrite both sides of the equation in terms of the same base.
???(5^2)^x=5^4???
???5^{2x}=5^4???
Since the bases are equivalent, that means the exponents must also be equivalent in order for the equation to be true.
???2x=4???
???x=2???
Therefore, the value of the original expression is ???2???, or
???\frac{\log{625}}{\log{25}}=2???
We can also solve other kinds of exponential equations using logs and the change of base formula.
Example
Use logs to solve the equation.
???10\cdot5^{2x}=300???
In problems like this, we have an equation, and we need to solve for the variable, ???x???, which means we’re trying to get ???x??? by itself on one side of the equation. In this particular example, we can start by dividing through by ???10???.
???10\cdot5^{2x}=300???
???5^{2x}=30???
Now we can use the general rule for logs to change this into a logarithmic equation.
???\log_5{30}=2x???
We’ll apply the change of base formula,
???2x=\frac{\log{30}}{\log{5}}???
And then we can solve for the variable.
???x=\frac{\log{30}}{2\log{5}}???
This is the exact value of the variable, but we can also use a calculator to find the decimal value.
???x\approx1.0566???