How do you solve ${log}_{10} y = (\frac{1}{4}) {log}_{10} (16) + (\frac{1}{2}) {log}_{10} (25)$ $log_10y = (1/4)log_10(16) + (1/2)log_10(25)$ ?

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How do you solve ${log}_{10} y = (\frac{1}{4}) {log}_{10} (16) + (\frac{1}{2}) {log}_{10} (25)$ $log_10y = (1/4)log_10(16) + (1/2)log_10(25)$ ?

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Answer 1 · 2015-12-21T07:12:03

$y = 10$ $y = 10$

Explanation:

${log}_{10} (y) = (\frac{1}{4}) {log}_{10} (16) + (\frac{1}{2}) {log}_{10} (25)$ $log_10 (y) = (1/4) log_10 (16) + (1/2) log_10(25)$

Rules which can be used here.

$n \cdot log (a) = log (a^{n})$ $n*log(a) = log(a^n)$
$log (a) + log (b) = log (a b)$ $log(a) + log(b) = log(ab)$
If $log (a) = log (b)$ $log(a) = log(b)$ then $a = b$ $a=b$

${log}_{10} (y) = {log}_{10} {(16)}^{\frac{1}{4}} + {log}_{10} {(25)}^{\frac{1}{2}}$ $log_10 (y) = log_10 (16)^(1/4) + log_10 (25)^(1/2)$ By rule 1.
${log}_{10} (y) = {log}_{10} (2) + {log}_{10} (5)$ $log_10 (y) = log_10 (2) + log_10 (5)$ since $a^{\frac{1}{n}} = \sqrt[n]{a}$ $a^(1/n) = root(n)a$
${log}_{10} (y) = {log}_{10} (2 \cdot 5)$ $log_10 (y) = log_10 (2*5)$ By rule 2.
${log}_{10} (y) = {log}_{10} (10)$ $log_10 (y) = log_10 (10)$

$y = 10$ $y = 10$ By Rule 3.

$y = 10$ $y = 10$ is the final answer.

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How do you solve ${log}_{10} y = (\frac{1}{4}) {log}_{10} (16) + (\frac{1}{2}) {log}_{10} (25)$ $log_10y = (1/4)log_10(16) + (1/2)log_10(25)$ ?

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How do you solve ${log}_{10} y = (\frac{1}{4}) {log}_{10} (16) + (\frac{1}{2}) {log}_{10} (25)$ $log_10y = (1/4)log_10(16) + (1/2)log_10(25)$ ?