How do you solve ${log}_{2} x - {log}_{8} x = 4$ $log_2 x - log_8 x = 4$ ?

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Answer 1 · 2016-04-07T02:58:52

You must first put in the same base. This can be started out by using the log rule ${log}_{a} n = \frac{log n}{log a}$ $log_an = logn/loga$

$\frac{log x}{log 2} - \frac{log x}{log 8} = 4$ $logx/log2 - logx/log8 = 4$

Now, rewrite the terms in the denominator by using the rule ${log a}^{n} = n log a$ $loga^n = nloga$

$\frac{log x}{1 log 2} - \frac{log x}{{log 2}^{3}} = 4$ $logx/(1log2) - logx/(log2^3) = 4$

$\frac{log x}{1 log 2} - \frac{log x}{3 log 2} = 4$ $logx/(1log2) - logx/(3log2) = 4$

Place on an equal denominator.

$\frac{3 log x}{3 log 2} - \frac{log x}{3 log 2} = 4$ $(3logx)/(3log2) - logx/(3log2) = 4$

${log}_{8} (x^{3}) - {log}_{8} (x) = 4$ $log_8(x^3) - log_8(x) = 4$

Now, you must use the rule ${log}_{a} n - {log}_{a} m = {log}_{a} (\frac{n}{m})$ $log_an - log_am = log_a(n/m)$ .

${log}_{8} (\frac{x^{3}}{x}) = 4$ $log_8(x^3/x) = 4$

Convert to exponential form:

$x^{2} = 8^{4}$ $x^2 = 8^4$

$x^{2} = 4096$ $x^2 = 4096$

$x = 64$ $x = 64$

Checking the solution in the equation, we find that it works. Thus our solution set is ${x = 64}$ ${x = 64}$

Hopefully this helps!

How do you solve log2x−log8x=4log_2 x - log_8 x = 4?