Change of Base: Make Different Logarithms Work Together

One logarithm, written on a different scale

Suppose you want $\log_5 12$, but your calculator offers only $\log_{10}$ and $\ln$. You can use either. Changing base rewrites the same number as a ratio:

The original argument must satisfy $a>0$; the original base needs $b>0$ and $b\ne1$. The new base must also satisfy $c>0$, $c\ne1$. Because $b\ne1$, the denominator $\log_c b$ is not zero.

Here is why the formula works. Let $y=\log_b a$, so $b^y=a$. Taking a base-$c$ logarithm gives $y\log_c b=\log_c a$. Dividing by the nonzero $\log_c b$ produces the formula. The numerator belongs to the argument; the denominator belongs to the original base.

The new base is a choice, not an extra unknown. Choose a convenient one: base 2 for powers of 2, base 3 for powers of 3, or a calculator’s available logarithm for a decimal estimate. If both numerator and denominator are changed consistently, the ratio stays the same.

In the activity, keep the original argument and base fixed while changing the calculation base. Predict whether the result will change before revealing it. A rounded calculator display is an approximation; an exact answer such as $\tfrac23$ should remain a fraction when the powers make that possible.

Swap the base and argument carefully

Changing base reveals a useful pair. For $a,b>0$, with neither equal to 1,

The two values are reciprocals, so

Here $\ln$ means the logarithm to base $e$; its exact numerical value is not needed for the cancellation. Both $\ln a$ and $\ln b$ must be nonzero. This explains why the reciprocal rule needs both numbers to differ from 1.

The restriction is easy to overlook. $\log_2 1=0$ is perfectly valid. But you cannot rewrite it as $1/\log_1 2$, because the proposed logarithm has the invalid base 1. A useful identity does not make an undefined expression legal.

Likewise, compare the parentheses in $(\log_a b)^{-1}$ and $\log_a(b^{-1})$. The first is the reciprocal of a logarithm; the second is the logarithm of a reciprocal. The power law makes the second equal to $-\log_a b$, which is generally different from $1/\log_a b$.

When a denominator contains one logarithm, swapping its base and argument can make the next step simpler. If the denominator contains a sum, pause. You must deal with the whole denominator before applying the reciprocal rule; it does not distribute over addition.

Find the beginning and end of a chain

A product can look complicated because each logarithm has a different base. Before evaluating anything, look for a link: does the argument of one logarithm become the base of the next?

The middle quantity disappears because the same nonzero factor $\ln a$ occurs in the numerator and denominator. This is ordinary multiplication of fractions after change of base, not crossing out matching printed numbers. The conditions are $a,b>0$, $a,b\ne1$, and $c>0$.

A longer complete chain works in the same way. In $\log_2 3\cdot\log_3 4\cdot\log_4 5$, every middle link matches. The surviving original base is 2, and the surviving final argument is 5. The result is $\log_2 5$, which need not be an integer.

Be careful with the dots in a long expression. They must stand for the entire specified chain. A missing link leaves an uncancelled factor. Also, a sum of changing-base logs is not this product and cannot be simplified by the same cancellation.

Use the activity to change the last integer. Try a final value that is a power of 2, then one that is not. The structure still simplifies even when the exact answer remains a logarithm. Write the first base and final argument before doing the algebra; they are useful checks on your result.

See the symmetry inside two awkward powers

JEE extension. You do not always need a decimal approximation to compare large-looking powers. The identity

reveals a symmetry between $a$ and $c$. Its conditions are $a,c>0$, with $b>0$ and $b\ne1$. Notice that $a$ or $c$ may equal 1 here: neither is required to be a logarithm base in the displayed formula.

To see the reason, write $a=e^{\ln a}$ and use change of base. The left side becomes

Interchanging $a$ and $c$ does not change the product $\ln a\,\ln c$, so the right side has exactly the same value. The positivity conditions make each real power and logarithm in this argument legal.

This is especially helpful when the two expressions are being subtracted, divided or compared. Look for the symmetry before reaching for a calculator. It can turn two difficult calculations into a single structural observation.

The identity also works alongside the inverse rule. For $4^{\log_2 3}$, write $4=2^2$, then $4^{\log_2 3}=(2^{\log_2 3})^2=9$. Alternatively, interchange to get $3^{\log_2 4}=3^2$. Two different legal routes give the same exact result.

Keep the exponent brackets visible. $a^{\log_b c}$ is one positive power; it is not a product of $a$, a logarithm and $c$.

Treat the whole denominator as one object

JEE extension. A cyclic expression repeats the same pattern while the letters rotate. Consider a denominator such as $1+\log_a b+\log_a c$. The extra 1 is useful: since $1=\log_a a$, the whole denominator becomes $\log_a(abc)$.

Only now should you take its reciprocal:

You have changed three different bases into a single common base, $abc$. Rotating the letters produces $\log_{abc}b$ and $\log_{abc}c$. Their sum is therefore $\log_{abc}(abc)=1$.

The domain matters twice. The original bases $a,b,c$ must be positive and different from 1. Also require $abc\ne1$: if the product were 1, every original denominator would become $\log_a1=0$, making division impossible. This same condition makes the new common base $abc$ legal.

Do not split a reciprocal across addition. In general, $1/(u+v)\ne1/u+1/v$. For instance, $1/(2+3)=1/5$, whereas $1/2+1/3=5/6$. The correct simplification works by combining the denominator first, not by reciprocating each piece.

The technique is a useful pattern-recognition exercise: make each denominator one logarithm, swap base and argument, then combine the resulting same-base sum. If a condition is missing, establish it before using the pattern.