Using logarithms to simplify expressions

Let's now learn how to create and solve exercises using logarithms to simplify expressions.

The following quotient might be solved perfectly with a calculator and patience:

$$$\dfrac{7^{15}\cdot 5^{\frac{2}{3}}}{9^{\frac{3}{4}}\cdot3^{37}}$$$

But when dealing with big numbers, like an exponent $$15$$ and another of $$37$$, we can simplify by appliying logarithms.

When we use logarithms, a product of numbers becomes a sum, a quotient is transformed into subtraction and a power into a product.

That's why the previous example can be solved applying decimal logarithms:

$$$log\dfrac{7^{15}\cdot 5^{\frac{2}{3}}}{9^{\frac{3}{4}}\cdot3^{37}}=\Big(log7^{15}+log5^{\frac{2}{3}}\Big)-\Big(log9^{\frac{3}{4}}+log3^{37}\Big)=$$$ $$$=(15\cdot log7+\dfrac{2}{3}\cdot log5)-(\dfrac{3}{4}\cdot log9+37\cdot log3)\simeq$$$ $$$\simeq(15\cdot0,845 +\dfrac{2}{3}\cdot0,699)-(\dfrac{3}{4}\cdot0,954+37\cdot0,477)\simeq$$$ $$$\simeq12,675+0,466-0,716-17,649\simeq -5,224$$$

It is necessary to remember that this result is the exponent that must be raised to $$10$$ (since we have to use decimal logarithms) to obtain the initial quotient of powers, so that:

$$$\dfrac{7^{15}\cdot 5^{\frac{2}{3}}}{9^{\frac{3}{4}}\cdot3^{37}}\simeq 10^{-5,224}\simeq 5,97\cdot 10^{-6}$$$

The following expression can also be simplified using logarithms:


Specifically, we can apply the logarithm in base $$x$$, then it is possible to express the number in the same base and simplify the quotient:

$$$log_x \dfrac{x}{2x\cdot\sqrt[3]{x}}=log_x x - \Big(log_x 2x - log_x x^{\frac{1}{3}}\Big)=$$$ $$$=1-2-\dfrac{1}{3}=-1-\dfrac{1}{3}=-\dfrac{3}{3}-\dfrac{1}{3}=-\dfrac{4}{3}$$$

The obtained result will be the number that must be raised to $$x$$, since we have used the unknown as a base of the logarithm, so that:


This ability to simplify the calculations of the logarithms also can be applied to equations. It can be useful, for example, when the unknown is written as an exponent.

In the equation $$2^x=10$$

If logarithms are added on both sides of the equality we obtain

$$log2^x=log10 \Rightarrow x\cdot log2=log10 \Rightarrow x=\dfrac{1}{log2}\simeq\dfrac{1}{0,301}\simeq 3,322$$

Using the property of the power of a logarithm it turns out to be easy to isolate $$x$$ in an expression like the previous one, and that is why it is so useful.