Question

Write the expression as a single logarithm 7log(x) y + 6 log(b) x

Answer 1

Answer: $log(x)^{7y}+log(b)^{6x}=log(x^{7y}b^{6x}).$

Step-by-step explanation:

Given logarithm expression

$7log(x)^y + 6 log(b)^x.$

We need to write it as a single logarithm expression.

First we need to apply exponent rule of logs to remove 7 and 6 in front of logs.

$n log(p) = log(p)^n$

$7log(x)^y + 6 log(b)^x.= log(x)^{7y}+log(b)^{6x}$.

Now, we need to apply product rule of logs $log(p)+log(q) = log(pq)$.

$log(x)^{7y}+log(b)^{6x}=log(x^{7y}b^{6x}).$

Therefore, final answer is $log(x)^{7y}+log(b)^{6x}=log(x^{7y}b^{6x}).$