2021-08-02

Exponential and Logarithmic Equations solve the equation. Find the exact solution if possible. otherwise, use a calculator to approximate to two decimals.
${\mathrm{log}}_{x}\left(x+5\right)-{\mathrm{log}}_{8}\left(x-2\right)=1$

Macsen Nixon

Step 1
Law of logarithm:
Consider m to be a positive number and $m\ne q1.$
Again consider M and M to be any real numbers with $M>0$ and $N>0.$
The difference of logarithms of two numbers is equal to the logarithm of quotient of two numbers as,
${\mathrm{log}}_{m}M-{\mathrm{log}}_{m}N={\mathrm{log}}_{m}\left(\frac{M}{N}\right)$
The logarithm function with base m is denoted by ${\mathrm{log}}_{m}$ can be defined as,
${\mathrm{log}}_{m}M=y$
$M={m}^{y}$
Step 2
The given logarithm equation is,
1) ${\mathrm{log}}_{8}\left(x+5\right)-{\mathrm{log}}_{8}\left(x-2\right)=1$
The above logarithm equation can be combined from the laws of logarithm as,
2) ${\mathrm{log}}_{8}\frac{\left(x+5\right)}{\left(x-2\right)}=1$
The equation (2) can be expressed as,
$\frac{\left(x+5\right)}{\left(x-2\right)}={8}^{1}$
$x+5=8\left(x-2\right)$
$x+5=8x-16$
$x-8x=-16-5$
Simplify above equation as,
$-7x=-21$
$7x=21$
$x=\frac{21}{7}$
$x=3$
Therefore, $x=3$ is the solution of equation ${\mathrm{log}}_{8}\left(x+5\right)-{\mathrm{log}}_{8}\left(x-2\right)=1.$
Conclusion:
Thus, the solution of the logarithm equation ${\mathrm{log}}_{8}\left(x+5\right)-{\mathrm{log}}_{8}\left(x-2\right)=1$ is 3.

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