Unfolding Change of Base Formula Mystery

The property of logarithm to change its base and retain its equality is really amazing. To unveil this mystery we need to know how it works. Knowing this property makes us capable of solving hard problems in logarithmic equations.


Important properties to take note when dealing with logarithm,

\log(x^n)=n\log(x) \log_a(a^n)=n \log_b(1)=0

General formula:



\log_25=\frac{\log5}{\log2} with the same base of 10.


Worked Problem 1: 2013 MMC Elimination

Write (\log_37)(\log_43)(\log_76) as a single logarithm.


Using the change of base formula we will be able to work with this problem easily.

$late \log37=\frac{\log7}{\log3}$




Observe that we can cancel same terms leaving

\displaystyle\frac{\log6}{\log4} but this can be simplified further using reversed change of base formula,



Worked Problem 2:

Solve for x: \log_4(x)+\log_{16}(4x^2)= \displaystyle\frac{5}{2}


In solving logarithmic equation we need to make sure first that the bases of expressions involved are the same. The base of the first expression is 4 and the other one is 2. So we need to find ways to make those bases the same. Right! We use change of base formula.

We need to express \log_4(x) to base 2 or \log_{16}(4x^2) to base 4. It’s up to your choice.

Mine is the second one.

Expressing \log_{16}(4x^2) to base 4,


Going back to original equation,






\log_4(x^2\cdot 4x^2)=5






Worked Problem 3:

Solve for x: \log_{x^2}(4x-4)+\log_{(4x-4)}(x^2)=2



Using change base formula,


By cross multiplication,

\displaystyle\frac{(\log(4x-4))^2+(\log(x^2))^2}{\log(4x-4)\cdot \log(x^2)}=2

(\log(4x-4))^2+(\log(x^2))^2=2\cdot \log(4x-4)\cdot \log(x^2)

Let a=\log(4x-4) and b=\log(x^2)

The equation is simplified as:




Squaring both sides,




Plugging back the values of a and b,


4x-4=x^2 (Note! They are equal by comparison, not by rule of logarithm.)





Worked Problem 4:

\displaystyle\frac{1}{log_2{1000!}} +\displaystyle\frac{1}{log_3{1000!}}+ \displaystyle\frac{1}{log_4{1000!}}+\dots+\displaystyle\frac{1}{log_{1000}{1000!}}


Let’s start with \displaystyle\frac{1}{log_2{1000!}}

Using change of base,


\to \displaystyle\frac{\log2}{\log1000!}

This will happen the same in other terms. Applying change of base to all expressions we have,


Take note that the denominator is the same.

\displaystyle\frac{\log 2+\log 3+\log 4+\dots+\log 1000}{\log 1000!}

Using multiplication-addition law mentioned above in numerator,

\displaystyle\frac{\log(2\cdot 3 \cdot 4\cdot \dots \cdot 1000)}{\log1000!}



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