How do you solve $2 {log}_{4} x - {log}_{4} (x - 1) = 1$ $2log_4x-log_4(x-1)=1$ ?

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Answer 1 · 2016-10-12T13:24:33

$x = 2$ $x = 2$

Start by applying the rule $a log n = {log n}^{a}$ $alogn = logn^a$ .

${log}_{4} (x^{2}) - {log}_{4} (x - 1) = 1$ $log_4(x^2) - log_4(x- 1) = 1$

Now apply the difference rules of logarithms: ${log}_{a} (n) - {log}_{a} (m) = {log}_{a} (\frac{n}{m})$ $log_a(n) - log_a(m) = log_a(n/m)$ .

${log}_{4} (\frac{x^{2}}{x - 1}) = 1$ $log_4((x^2)/(x - 1)) = 1$

Convert to exponential form using the rule ${log}_{a} (n) = b \to a^{b} = n$ $log_a(n) = b -> a^b = n$

$\frac{x^{2}}{x - 1} = 4^{1}$ $(x^2)/(x - 1) = 4^1$

$\frac{x^{2}}{x - 1} = 4$ $x^2/(x - 1) = 4$

$x^{2} = 4 (x - 1)$ $x^2 = 4(x - 1)$

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

$x^{2} - 4 x + 4 = 0$ $x^2 - 4x + 4 = 0$

$(x - 2) (x - 2) = 0$ $(x - 2)(x - 2) = 0$

$x = 2$ $x = 2$

Checking in the original equation:

${log}_{4} (2^{2}) - {log}_{4} (2 - 1) =^{?} 1$ $log_4(2^2) - log_4(2 - 1) =^? 1$

${log}_{4} (4) - {log}_{4} (1) =^{?} 1$ $log_4(4) - log_4(1) =^?1$

$1 - 0 = 1$ $1 - 0 = 1$

The solution works!

Hopefully this helps!

How do you solve 2log4x−log4(x−1)=12log_4x-log_4(x-1)=1?