Logarithm addition with different bases

Question

How to solve the following? $$\log_2{8}+ \log_3{27}$$ Please advise whether my method below is correct or not? $$\begin{align*}\log_2{8}&=3 \\ \log_3{27}&=3 \end{align*}$$ Answer is: $3+3=6$.

Answer 1

Yes, you are correct. $\log_2{8}$ can be thought of as what integer $x$ will make $2^x = 8$ a true statement. The (only) solution to that statement is $x = 3$. Similarly, $\log_3{27}$ can be thought of as what integer $y$ will make $3^y = 27$ true. The (only) answer would be $y = 3$.

Thus, $$\log_2{8} + \log_3{27} = 3 + 3 = 6$$ as what you would expect.

Answer 2

Correct Answer:

$\log_2 8 + \log_3 27$

Log rule for different bases: $\log_b a = \dfrac {\log a}{\log b}$

So

$\log_2 8 + \log_3 27$

$\dfrac {\log 8}{\log 2} + \dfrac {\log 27}{\log 3}$ (cancel the logs)

$\dfrac 8 2 + \dfrac {27} 3$

$ 4 + 9 = 13$