# How do you solve log_5(5)+log_5(2x)+log_5(3)=98 ?

Jan 5, 2016

Step by step explanation is given below.

#### Explanation:

Before starting out on the problem let us quickly go through few rules.

$1. {\log}_{b} \left(a\right) = k \implies a = {b}^{k}$ this is used to covert logarithm into exponent form.
$2 , {\log}_{b} \left(b\right) = 1$
$3. {\log}_{b} \left(P\right) + {\log}_{b} \left(Q\right) = {\log}_{b} \left(P Q\right)$ This rule for condensing the log sum to product.

For our problem, these rules are enough. For the other type of logarithmic equations, please do spend some time learning all the basic rules.

${\log}_{5} \left(5\right) + {\log}_{5} \left(2 x\right) + {\log}_{5} \left(3\right) = 98$

$1 + {\log}_{5} \left(\left(2 x\right) \left(3\right)\right) = 98$

Subtract $1$ from both the sides

${\log}_{5} \left(6 x\right) = 97$

$6 x = {5}^{97}$

Dividing by $6$ on both the sides to solve for $x$

$x = {5}^{97} / 6$ Answer