Can you help me with this logarithms equation for my test? ASAP

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Answer 1 · 2018-06-26T21:47:39

$x = 16$ $x = 16$

Use log laws:

${log}_{a} λ + {log}_{a} μ = {log}_{a} λ μ$ $log_a lamda + log_a mu = log_a lamda mu$

$\Rightarrow {log}_{4} (x (x - 12)) = 3$ $=> log_4(x(x-12) ) =3$

Use another log law:

$a^{{log}_{a} γ} = γ$ $a^(log_a gamma)= gamma$

Hence $4^{{log}_{4} (x (x - 12))} = 4^{3}$ $4^(log_4( x(x-12))) = 4^3$

$\Rightarrow x (x - 12) = 64$ $=> x(x-12) = 64$

$\Rightarrow x^{2} - 12 x - 64 = 0$ $=> x^2 -12x - 64 = 0$

$x = {- 4, 16}$ $x = {-4, 16 }$

But $log( "negative" ) notin RR$

Hence Reject $x = -4 ->x = 16$ is our final solution