Question
Question: How do you solve \(\log (5x + 2) = \log (2x - 5)\)?...
How do you solve log(5x+2)=log(2x−5)?
Solution
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=3−7 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) ⇒loga=logb
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 ⇒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:-
33x=3−7 ⇒x=3−7
Therefore, we can conclude that x=3−7.
Solution of log(5x+2)=log(2x−5)isx=3−7.
Now, to check whether this solution is correct or not, we will put x=3−7in log(5x+2)=log(2x−5).
So,
log(5x+2)=log(2x−5) ⇒log(5×3−7+2)=log(2×3−7−5) ⇒log(3−35+2)=log(3−14−5) ⇒log(3−35+6)=log(3−14−15) ⇒log(3−29)=log(3−29) ∴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.