How To Get Rid Of Logarithms

Nothing messes up an equation quite like logarithms. They are cumbersome, difficult to manipulate and a little mysterious for some people. Luckily, there's an easy way to rid your equation of these pesky mathematical expressions. All you have to do is remember that a logarithm is the inverse of an exponent. Although the base of a logarithm can be any number, the most common bases used in science are 10 and e, which is an irrational number known as Euler's number. To distinguish them, mathematicians use "log" when the base is 10 and "ln" when the base is e.

The concept of a logarithm is simple, but it's a little difficult to put into words. A logarithm is the number of times you have to multiply a number by itself to get another number. Another way to say it is that a logarithm is the power to which a certain number – called the base – must be raised to get another number. The power is called the argument of the logarithm.

For example, log₈2 = 64 simply means that raising 8 to the power of 2 gives 64. In the equation log x = 100, the base is understood to be 10, and you can easily solve for the argument, x because it answers the question, "10 raised to what power equals 100?" The answer is 2.

A logarithm is the inverse of an exponent. The equation log x = 100 is another way of writing 10_^x_ = 100. This relationship makes it possible to remove logarithms from an equation by raising both sides to the same exponent as the base of the logarithm. If the equation contains more than one logarithm, they must have the same base for this to work.

In the simplest case, the logarithm of an unknown number equals another number:

\(\log x = y\)

Raise both sides to exponents of 10, and you get

\(10^ {\log x} = 10^y\)

Since 10^{(log x)} is simply x, the equation becomes

\(x = 10^y\)

When all the terms in the equation are logarithms, raising both sides to an exponent produces a standard algebraic expression. For example, raise

\(\log (x^2 – 1) = \log (x + 1)\)

to a power of 10 and you get:

\(x^2 – 1 = x + 1\)

which simplifies to

\(x^2 – x – 2 = 0.\)

The solutions are x = −2; x = 1.

In equations that contain a mixture of logarithms and other algebraic terms, it's important to collect all the logarithms on one side of the equation. You can then add or subtract terms. According to the law of logarithms, the following is true:

\(\log x + \log y = \log(xy) \
\,\
\log x – \log y = \log \bigg(\frac{x}{y}\bigg)\)

Here's a procedure for solving an equation with mixed terms:

Start with the equation: For example

\(\log x = \log (x – 2) + 3\)

Rearrange the terms:

\(\log x – \log (x – 2) = 3\)

Apply the law of logarithms:

\(\log \bigg(\frac{x}{x-2}\bigg) = 3\)

Raise both sides to a power of 10:

\(\bigg(\frac{x}{x-2}\bigg) = 10^3\)

Solve for x:

\(\bigg(\frac{x}{x-2}\bigg) = 10^3 \
x = 1000x – 2000 \
-999x = -2000 \
x = \frac{2000}{999}=2.002\)