Question

If log2

x + log4

x + log16x = 21/4, these x is

equal to​

Answer 1

∴ $x$ is equal to $10^{\frac{7}{3} (\frac{3}{4}-log2)}$

Step-by-step explanation:

Given;

$log2x+log4x+log16x=\frac{21}{4}$

⇒$log(2x\times4x\times16x)=\frac{21}{4}$  (∵$loga+logb=log(a\times b)$ )

⇒$log(128x^{3})=\frac{21}{4}$

⇒$log128+logx^{3}=\frac{21}{4}$

⇒$log2^{7} +3logx=\frac{21}{4}$  (∵$logx^{a} =alogx$)

⇒$7log2 +3logx=\frac{21}{4}$

⇒$3logx=\frac{21}{4}-7log2$

⇒$logx=\frac{7}{3} (\frac{3}{4}-log2)$

∴$x=10^{\frac{7}{3} (\frac{3}{4}-log2)}$

So, $x$ is equal to $10^{\frac{7}{3} (\frac{3}{4}-log2)}$