Question

How do you solve $\log x+\log 7=\log 37$?

Answer 1

Hint: We solve the given equation using the different identity formulas of logarithm like $\log a+\log b=\log \left( ab \right)$, \[\log a=\log b\Rightarrow a=b\]. The main step would be to eliminate the logarithm function and keep only the quadratic equation of $x$. We solve the equation with the help of division.

Complete step by step answer:
We take the logarithmic identity for the given equation $\log x+\log 7=\log 37$ to find the solution for x. We have $\log a+\log b=\log \left( ab \right)$.We operate the addition part in the left-hand side of $\log x+\log 7=\log 37$.
$\log x+\log 7=\log 37 \\
\Rightarrow \log 7x=\log 37 \\ $
We know \[\log a=\log b\Rightarrow a=b\]. Applying the rule in case of $\log 7x=\log 37$, we get
$\log 7x=\log 37 \\
\Rightarrow 7x=37 \\ $
Now we have a linear equation of x. We need to solve it. We divide both sides with 7 and get $x=\dfrac{37}{7}$.

Therefore, the solution of $\log x+\log 7=\log 37$ is $x=\dfrac{37}{7}$.

Note: In case the base is not mentioned then the general solution for the base for logarithm is 10. But the base of $e$ is fixed for $\ln $. We also need to remember that for logarithm function there has to be a domain constraint. For any ${{\log }_{b}}a$, $a>0$. This means for $\log x+\log 7=\log 37$, $x>0$.