Question

How do you solve $\log (5x + 2) = \log (2x - 5)$?

Answer 1

Hint:Equate the argument of logs on both sides. To solve this question, the first thing we will do is to equate the arguments of both the log. So, $5x + 2 = 2x - 5$. Now, we will solve this equation for x as it only has one single variable. This step will give us the value of $x = \dfrac{{ - 7}}{3}$ which will be the final solution of $\log (5x + 2) = \log (2x - 5)$.

Complete step by step solution:
The given equation we have is $\log (5x + 2) = \log (2x - 5)$
Now, let’s consider
$a = 5x + 2$
And
$b = 2x - 5$
So, we can rewrite the given equation as:-
$
\log (5x + 2) = \log (2x - 5) \\
\Rightarrow \log a = \log b \\
$
We can now see that, both the LHS and RHS side have logs on both sides.
And since log is the only term on both the sides a must be equal to b.
Hence, if we equate the two terms we will get:-
$
a = b \\
\Rightarrow 5x + 2 = 2x - 5 \\
$
Now, subtracting 2 both the sides we will get:-
$
5x + 2 = 2x - 5 \\
5x + 2 - 2 = 2x - 5 - 2 \\
5x = 2x - 7 \\
$
To simplify further, we will subtract 2x on both the sides, we will get:-
$
5x - 2x = 2x - 2x - 7 \\
$
Now, we will divide each side by 3, we will get:-
$
\dfrac{{3x}}{3} = \dfrac{{ - 7}}{3} \\
\Rightarrow x = \dfrac{{ - 7}}{3} \\
$
Therefore, we can conclude that $x = \dfrac{{ - 7}}{3}$.
Solution of $\log (5x + 2) = \log (2x - 5)$is$x = \dfrac{{ - 7}}{3}$.
Now, to check whether this solution is correct or not, we will put $x = \dfrac{{ - 7}}{3}$in $\log (5x + 2) = \log (2x - 5)$.
So,
$
\log (5x + 2) = \log (2x - 5) \\
\Rightarrow \log \left( {5 \times \dfrac{{ - 7}}{3} + 2} \right) = \log \left( {2 \times \dfrac{{ - 7}}{3} -
5} \right) \\
\Rightarrow \log \left( {\dfrac{{ - 35}}{3} + 2} \right) = \log \left( {\dfrac{{ - 14}}{3} - 5} \right)
\\
\Rightarrow \log \left( {\dfrac{{ - 35 + 6}}{3}} \right) = \log \left( {\dfrac{{ - 14 - 15}}{3}} \right)
\\
\Rightarrow \log \left( {\dfrac{{ - 29}}{3}} \right) = \log \left( {\dfrac{{ - 29}}{3}} \right) \\
\therefore LHS = RHS \\
$
Hence, our answer is correct.

Note: Arguments of both the logs are equated only because the question had only log terms. If any one side had 1 or more than 1 term extra, then we would not have done this step. In those cases, we would have simplified the question more to get a more step and solve the question from there.