If-log-4-log-2-x-log-2-log-4-x-2-then-log-5-x-x-5-

Question Number 133615 by benjo_mathlover last updated on 23/Feb/21

If \log_{4} (\log_{2} x) + \log_{2} (\log_{4} x) = 2

then \log_{5} \sqrt{x + \sqrt{x} + 5} = ?

Answered by TheSupreme last updated on 23/Feb/21

{l o g}_{4} (x) = \frac{{l o g}_{2} x}{{l o g}_{2} 4} = \frac{1}{2} {l o g}_{2} x

\frac{1}{2} {l o g}_{2} ({l o g}_{2} (x)) + {l o g}_{2} (\frac{1}{2} {l o g}_{2} (x)) = 2

{l o g}_{2} [\frac{1}{2} {l o g}_{2} {(x)}^{\frac{3}{2}}] = 2

\frac{1}{2} {l o g}^{\frac{3}{2}} (x) = 4

{l o g}^{\frac{3}{2}} (x) = 8 \to {l o g}_{2} (x) = 2^{3 \times \frac{2}{3}} = 4

x = 2^{4} = 16

{l o g}_{5} (\sqrt{16 + \sqrt{16} + 5}) = {l o g}_{5} (25) = 2

Commented by JDamian last updated on 24/Feb/21

\sqrt{16 + \sqrt{16} + 5} \neq 25

\sqrt{16 + \sqrt{16} + 5} = \sqrt{25} = 5

Answered by liberty last updated on 25/Feb/21

let \log_{4} x = u \Rightarrow \log_{2} x = 2 u

\Leftrightarrow \log_{4} (2 u) + \log_{2} (u) = 2

\Rightarrow \log_{2} (2 u) + 2 \log_{2} (u) = 4

\Rightarrow \log_{2} (2 u \times u^{2}) = \log_{2} (16)

\Rightarrow u^{3} = 8 \Rightarrow u = 2

\log_{2} (x) = 4 \Rightarrow x = 16 then

\log_{5} \sqrt{16 + \sqrt{16} + 5} = \log_{5} \sqrt{25} = 1

Commented by JDamian last updated on 24/Feb/21

\log_{5} \sqrt{25} \neq 2

\log_{5} \sqrt{25} = \log_{5} 5 = 1

Facebook Tweet Pin