# Logarithmic equations

A mathematical equation which contains logarithmic functions as terms is called a logarithmic equation.

## Introduction

The logarithmic terms are formed to express quantities mathematically. The expression of one or more connected log terms represents a quantity and the same quantity can also be expressed by different one or more connected log terms.

Then, the two logarithmic expressions are equal and the mathematical equation is completely in terms of log functions. Therefore, the mathematical equation is called as a logarithmic equation.

### Example

$3$ is a number. It can be expressed in logarithmic form.

$\log_{2}{(8)} = 3$

The quantity $3$ is equal to $\log_{2}{8}$. So, it is a basic mathematical equation. It can also be written in the following way as well.

$\implies \log_{2}{(8)}-3 = 0$

In this equation, $\log_{2}{8}$ is a logarithmic term. So, the mathematical equation is called as a logarithmic equation.

Now, forget about this example temporarily and subtract $\log_{5}{(25)}$ from $\log_{4}{(1024)}$.

$\log_{4}{(1024)}-\log_{5}{(25)}$

Now, simplify this log expression.

$=\,\,\, \log_{4}{(4^5)}-\log_{5}{(5^2)}$

As per power rule of logarithms, the two terms in the expression can be simplified.

$=\,\,\, 5\log_{4}{(4)}-2\log_{5}{(5)}$

The logarithm of base is always one as per log of base rule.

$=\,\,\, 5 \times 1-2 \times 1$

$=\,\,\, 5-2$

$=\,\,\, 3$

Therefore, $\log_{4}{(1024)}-\log_{5}{(25)} \,=\, 3$

But, $\log_{2}{(8)} = 3$ is also true. So, the expression $\log_{2}{(8)}$ can be equal to the logarithmic expression $\log_{4}{(1024)}-\log_{5}{(25)}$

$\,\,\, \therefore \,\,\,\,\,\, \log_{4}{(1024)}-\log_{5}{(25)}$ $\,=\,$ $\log_{2}{(8)}$

It is a mathematical equation completely in terms of logarithms. So, it is also called as a logarithmic equation. Thus, the logarithmic equations are formed in mathematics.

### Examples

Observe the following few more equations to have some basic knowledge about log equations.

$(1)\,\,\,\,\,\,$ $\log_{3}{[5+4\log_{3}{(x-1)}]}$ $\,=\, 2$

$(2)\,\,\,\,\,\,$ $\log{\Bigg[\dfrac{1}{2^y+y-1}\Bigg]}$ $\,=\,$ $y{(\log{5}-1)}$

$(3)\,\,\,\,\,\,$ $\log_{2a}{2}$ $+$ $\log_{a}{2}$ $+$ $\dfrac{3}{2}$ $\,=\, 0$

Now, let’s learn solving logarithmic equations from example problems.

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