Write the expression as a single logarithm 7log(x) y + 6 log(b) x
1 answer:
Answer: ![log(x)^{7y}+log(b)^{6x}=log(x^{7y}b^{6x}).](https://tex.z-dn.net/?f=log%28x%29%5E%7B7y%7D%2Blog%28b%29%5E%7B6x%7D%3Dlog%28x%5E%7B7y%7Db%5E%7B6x%7D%29.)
Step-by-step explanation:
Given logarithm expression
![7log(x)^y + 6 log(b)^x.](https://tex.z-dn.net/?f=7log%28x%29%5Ey%20%2B%206%20log%28b%29%5Ex.)
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](https://tex.z-dn.net/?f=n%20log%28p%29%20%3D%20log%28p%29%5En)
.
Now, we need to apply product rule of logs
.
![log(x)^{7y}+log(b)^{6x}=log(x^{7y}b^{6x}).](https://tex.z-dn.net/?f=log%28x%29%5E%7B7y%7D%2Blog%28b%29%5E%7B6x%7D%3Dlog%28x%5E%7B7y%7Db%5E%7B6x%7D%29.)
Therefore, final answer is ![log(x)^{7y}+log(b)^{6x}=log(x^{7y}b^{6x}).](https://tex.z-dn.net/?f=log%28x%29%5E%7B7y%7D%2Blog%28b%29%5E%7B6x%7D%3Dlog%28x%5E%7B7y%7Db%5E%7B6x%7D%29.)
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