Question

if log(4x-3) + log(4x+3) = 4log2 then, find the valve of log 8x​

Answer 1

Answer:

$log(4x - 3) + log(4x + 3) = 4log2 \\ \\ \therefore \: log \{(4x - 3) \times (4x + 3) \} = log2^{4} \\ \\ \therefore \: log \{(4x)^{2} - (3) ^{2} \} = log16 \\ \\ \therefore \: 16x^{2} -9 = 16 \\ \\ \therefore \: 16x^{2} = 16 + 9 \\ \\ \therefore \: 16x^{2} = 25\\ \\ \therefore \: x^{2} = \frac{25}{16} \\ \\ \huge \red {\boxed{\implies \: x = \frac{5}{4} }} \\ \\ now \\ log \: 8x = log \: \{8 \times \frac{5}{4} \} \\ = log \: \{2 \times 5 \} \\ = log \: 10 \\ = 1 \\ thus \\ \huge \purple { \boxed{log \: 8x =1}}$