Question

log x base4+log x base2= 6 find x

Answer 1

$\underline{ \large{ \text{Answer}}} : \\ \\ \text{Given that,} \\ \\ log_4{x}+ log_2{x} = 6 \\ \\ \to \frac{logx}{log4} + \frac{logx}{log2} = 6 \: \: \lbrack \because log_b{a} = \frac{log_e{a}}{log_e{b}} \rbrack\\ \\ \to logx \: \biggr( \frac{1}{log4} + \frac{1}{log2} \biggr) = 6 \\ \\ \to logx \: \biggr( \frac{1}{2log2} + \frac{1}{log2} \biggr) = 6$

$\to logx \: \biggr( \frac{1 + 2}{2log2} \biggr) = 6 \\ \\ \to logx \: \biggr( \frac{3}{2log2} \biggr) = 6 \\ \\ \to logx = \frac{12log2}{3} \\ \\ \to logx = 4log2 \\ \\ \to logx = log {2}^{4} \: \lbrack \because b \: loga = log{a}^{b} \rbrack \\ \\ \to logx = log16 \\ \\ \therefore \boxed{ \bold{x = 16}}$

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Answer 2

Answer:

Step-by-step explanation: