1. ## Log Simplification?

I am stuck on something that shouldn't be particularly complicated.

I have logx(xlogyy5)

Should I get 1+logx5? This doesn't look right to me.

Instructions much appreciated.

2. ## Re: Log Simplification?

I think your approach is right. In case you find a discrepancy you may paste the same.

3. ## Re: Log Simplification?

Originally Posted by Quixotic
I am stuck on something that shouldn't be particularly complicated.

I have logx(xlogyy5)

Should I get 1+logx5? This doesn't look right to me.

Instructions much appreciated.
It's correct.

-Dan

4. ## Re: Log Simplification?

Hello, Quixotic!

$\displaystyle \text{I have: }\:\log_x(x\log_yy^5)$

$\displaystyle \text{Should I get: }\:1+\log_x5\,?$ . Yes!

$\displaystyle \log_5(x\log_yy^5) \;=\;\log_5(x\cdot 5\log_yy) \qquad\quad \text{Note that: }\log_yy =1$

. . . . . . . . . . $\displaystyle =\;\log_5(x\cdot 5\cdot 1)$

. . . . . . . . . . $\displaystyle =\;\log_5(5x)$

. . . . . . . . . . $\displaystyle =\;\log_55 + \log_5x \qquad\quad\text{Note that: }\log_55 = 1$

. . . . . . . . . . $\displaystyle =\; 1 + \log_5x$

5. ## Re: Log Simplification?

Thanks everyone!