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Question: How do you solve \(\log (5x + 2) = \log (2x - 5)\)?...

How do you solve log(5x+2)=log(2x5)\log (5x + 2) = \log (2x - 5)?

Explanation

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=2x55x + 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=73x = \dfrac{{ - 7}}{3} which will be the final solution of log(5x+2)=log(2x5)\log (5x + 2) = \log (2x - 5).

Complete step by step solution:
The given equation we have is log(5x+2)=log(2x5)\log (5x + 2) = \log (2x - 5)
Now, let’s consider
a=5x+2a = 5x + 2
And
b=2x5b = 2x - 5
So, we can rewrite the given equation as:-
log(5x+2)=log(2x5) loga=logb  \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 5x+2=2x5  a = b \\\ \Rightarrow 5x + 2 = 2x - 5 \\\
Now, subtracting 2 both the sides we will get:-
5x+2=2x5 5x+22=2x52 5x=2x7  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:-
5x2x=2x2x7  5x - 2x = 2x - 2x - 7 \\\
Now, we will divide each side by 3, we will get:-
3x3=73 x=73  \dfrac{{3x}}{3} = \dfrac{{ - 7}}{3} \\\ \Rightarrow x = \dfrac{{ - 7}}{3} \\\
Therefore, we can conclude that x=73x = \dfrac{{ - 7}}{3}.
Solution of log(5x+2)=log(2x5)\log (5x + 2) = \log (2x - 5)isx=73x = \dfrac{{ - 7}}{3}.
Now, to check whether this solution is correct or not, we will put x=73x = \dfrac{{ - 7}}{3}in log(5x+2)=log(2x5)\log (5x + 2) = \log (2x - 5).
So,
log(5x+2)=log(2x5) log(5×73+2)=log(2×735) log(353+2)=log(1435) log(35+63)=log(14153) log(293)=log(293) LHS=RHS  \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.