Question #c11ad

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Question #c11ad

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Answer 1 · 2016-12-26T19:53:20

see below

Explanation:

concepts applied

${log}_{a} (b^{c}) \Rightarrow c {log}_{a} (b)$ $\log_a(b^c)\rArrc\log_a(b)$
${log}_{a} (b) = x \leftrightarrow a^{x} = b$ $\log_a(b)=x\leftrightarrowa^x=b$

if referring to the log as base 10 log, ${log}_{10} x$ $log_10x$ :
$\frac{{log}_{10} (225)}{{log}_{10} (25)}$ $(\log_10(225))/(\log_10(25))$

then
numerator: ${log}_{10} (15^{2}) \Rightarrow 2 {log}_{10} (15)$ $\log_10(15^2)\rArr2\log_10(15)$
denominator: ${log}_{10} (5^{2}) \Rightarrow 2 {log}_{10} (5)$ $\log_10(5^2)\rArr2\log_10(5)$

becomes
$\frac{2 {log}_{10} (25)}{2 {log}_{10} (5)}$ $(2\log_10(25))/(2\log_10(5))$
the 2's cancel, which leaves you with $\frac{{log}_{10} (25)}{{log}_{10} (5)}$ $(\log_10(25))/(\log_10(5))$

simplify and you get your answer.

if referring to the log as natural log, ${log}_{e} x$ $log_ex$ or $ln x$ $lnx$ :
$\frac{{log}_{e} (225)}{{log}_{e} (25)} \Rightarrow \frac{ln (225)}{ln (25)}$ $(\log_e(225))/(\log_e(25))\rArr(\ln(225))/(\ln(25))$

then
numerator: $ln (15^{2}) \Rightarrow 2 ln (15)$ $\ln(15^2)\rArr2\ln(15)$
denominator: $ln (5^{2}) \Rightarrow 2 ln (5)$ $\ln(5^2)\rArr2\ln(5)$

becomes
$\frac{2 ln (25)}{2 ln (5)}$ $(2\ln(25))/(2\ln(5))$
the 2's cancel, which leaves you with $\frac{ln (25)}{ln (5)}$ $(\ln(25))/(\ln(5))$

simplify and you get your answer.

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Question #c11ad

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