How do you solve ${{3}^{x+1}}=15$ ?

Hint: We know that if the value of ${{a}^{x}}$ is equal to b then we can write x as ${{\log }_{a}}b$ for example we know that 3 to the power 2 is equal to 9. So we can write ${{\log }_{3}}9$ is equal to 2. Using this we can solve ${{3}^{x+1}}=15$ .

Complete step-by-step answer:
The given equation is ${{3}^{x+1}}=15$
We know that if ${{a}^{x}}$ is equal to b that implies x equal to ${{\log }_{a}}b$
So we can write x + 1 equal to ${{\log }_{3}}15$
Now x will be equal to ${{\log }_{3}}15-1$
We know that the value of ${{\log }_{a}}a$is equal to 1
So x = ${{\log }_{3}}15-{{\log }_{3}}3$
We know another formula of logarithm that $\log a-\log b$ is equal to $\log \dfrac{a}{b}$ where a and b are positive real number
So we can write ${{\log }_{3}}15-{{\log }_{3}}3$ is equal to ${{\log }_{3}}5$
So the value of x is equal to ${{\log }_{3}}5$
We can use the logarithm table to find the value of ${{\log }_{3}}5$ . ${{\log }_{3}}5$ is equal to $\dfrac{\ln 5}{\ln 3}$

Note: In the function ${{\log }_{a}}x$ we always take x a positive number , but the value of ${{\log }_{a}}x$ can be negative that depends up on the value of a and x. If the value of a and x are greater than 1 or the value of both number is less than 1 then value of ${{\log }_{a}}x$ will be positive in the other hand if one of a and b is greater than 1 and one is less than 1, then ${{\log }_{a}}x$ is negative. In ${{\log }_{a}}x$ , a can not be equal to 1.