How do you solve $4 {log}_{4} (x) - 9 {log}_{x} (4) = 0$ $4log_4 (x) - 9log_x (4) = 0$ ?

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How do you solve $4 {log}_{4} (x) - 9 {log}_{x} (4) = 0$ $4log_4 (x) - 9log_x (4) = 0$ ?

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Answer 1 · 2016-03-14T15:08:13

You must first simplify using the rule ${log}_{a} m = \frac{log m}{log a}$ $log_am = logm/loga$

Explanation:

First, we must use the rule $a log n = {log n}^{a}$ $alogn = logn^a$

${log}_{4} (x^{4}) - {log}_{x} (262144) = 0$ $log_4(x^4) - log_x(262144) = 0$

$\frac{{log x}^{4}}{log 4} - \frac{log 262144}{log x} = 0$ $(logx^4)/(log4) - (log262144)/logx = 0$

Place on a common denominator.

$\frac{log x ({log x}^{4})}{log 4 (log x)} - \frac{log 4 (log 262144)}{log x (log 4)} = 0$ $(logx(logx^4))/(log4(logx)) - (log4(log262144))/(logx(log4)) = 0$

Convert to exponential form.

$\frac{x^{5}}{1048576} = 10^{0}$ $x^5/1048576 = 10^0$

$x^{5} = 1048576$ $x^5 = 1048576$

$x = \sqrt[5]{1048576}$ $x = root(5)1048576$

$x = 16$ $x = 16$

Hopefully this helps!

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How do you solve $4 {log}_{4} (x) - 9 {log}_{x} (4) = 0$ $4log_4 (x) - 9log_x (4) = 0$ ?

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How do you solve $4 {log}_{4} (x) - 9 {log}_{x} (4) = 0$ $4log_4 (x) - 9log_x (4) = 0$ ?