Answer: [tex]log(x)^{7y}+log(b)^{6x}=log(x^{7y}b^{6x}).[/tex]
Step-by-step explanation:
Given logarithm expression
[tex]7log(x)^y + 6 log(b)^x.[/tex]
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.
[tex]n log(p) = log(p)^n[/tex]
[tex]7log(x)^y + 6 log(b)^x.= log(x)^{7y}+log(b)^{6x}[/tex].
Now, we need to apply product rule of logs [tex]log(p)+log(q) = log(pq)[/tex].
[tex]log(x)^{7y}+log(b)^{6x}=log(x^{7y}b^{6x}).[/tex]
Therefore, final answer is [tex]log(x)^{7y}+log(b)^{6x}=log(x^{7y}b^{6x}).[/tex]
