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

I have log_{x}(xlog_{y}y^{5})

Should I get 1+log_{x}5? This doesn't look right to me.

Instructions much appreciated.

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- Apr 8th 2013, 02:01 AMQuixoticLog Simplification?
I am stuck on something that shouldn't be particularly complicated.

I have log_{x}(xlog_{y}y^{5})

Should I get 1+log_{x}5? This doesn't look right to me.

Instructions much appreciated. - Apr 8th 2013, 02:14 AMibduttRe: Log Simplification?
I think your approach is right. In case you find a discrepancy you may paste the same.

- Apr 8th 2013, 04:34 AMtopsquarkRe: Log Simplification?
- Apr 8th 2013, 08:42 AMSorobanRe: Log Simplification?
Hello, Quixotic!

Quote:

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

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

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

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

- Apr 8th 2013, 12:59 PMQuixoticRe: Log Simplification?
Thanks everyone!