How do you solve $2log_4x-log_4(x-1)=1$ ?

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Answer 1 · 2016-10-12T13:24:33

$x = 2$

Start by applying the rule $alogn = logn^a$ .

$log_4(x^2) - log_4(x- 1) = 1$

Now apply the difference rules of logarithms: $log_a(n) - log_a(m) = log_a(n/m)$ .

$log_4((x^2)/(x - 1)) = 1$

Convert to exponential form using the rule $log_a(n) = b -> a^b = n$

$(x^2)/(x - 1) = 4^1$

$x^2/(x - 1) = 4$

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

$x^2 = 4x - 4$

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

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

$x = 2$

Checking in the original equation:

$log_4(2^2) - log_4(2 - 1) =^? 1$

$log_4(4) - log_4(1) =^?1$

$1 - 0 = 1$

The solution works!

Hopefully this helps!

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