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

$$\mathrm{If}\log {}_4\left(\log {}_2\mathrm{x}\right)+\log {}_2\left(\log {}_4\mathrm{x}\right)=2$$$$\mathrm{then}\log {}_5\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}+5}=?$$

Answered by TheSupreme last updated on 23/Feb/21

$${log}_4\left(x\right)=\frac{{log}_{2}x}{{log}_{2}4}=\frac{1}{2}{log}_{2}x$$$$\frac{1}{2}{log}_2\left({log}_2\left(x\right)\right)+{log}_2\left(\frac{1}{2}{log}_2\left(x\right)\right)=2$$$${log}_2\left[\frac{1}{2}{log}_2{\left(x\right)}^{\frac{3}{2}}\right]=2$$$$\frac{1}{2}{log}^{\frac{3}{2}}\left(x\right)=4$$$${log}^{\frac{3}{2}}\left(x\right)=8\to {log}_2\left(x\right)=2^{3\times \frac{2}{3}}=4$$$$x=2^4=16$$$${log}_5\left(\sqrt{16+\sqrt{16}+5}\right)={log}_5\left(25\right)=2$$

Commented by JDamian last updated on 24/Feb/21

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

Answered by liberty last updated on 25/Feb/21

$$\mathrm{let}\log {}_4\mathrm{x}=\mathrm{u}\Rightarrow \log {}_2\mathrm{x}=2\mathrm{u}$$$$\iff \log {}_4\left(2\mathrm{u}\right)+\log {}_2\left(\mathrm{u}\right)=2$$$$\Rightarrow \log {}_2\left(2\mathrm{u}\right)+2\log {}_2\left(\mathrm{u}\right)=4$$$$\Rightarrow \log {}_2\left(2\mathrm{u}\times {\mathrm{u}}^2\right)=\log {}_2\left(16\right)$$$$\Rightarrow {\mathrm{u}}^3=8\Rightarrow \mathrm{u}=2$$$$\log {}_2\left(\mathrm{x}\right)=4\Rightarrow \mathrm{x}=16\mathrm{then}$$$$\log {}_5\sqrt{16+\sqrt{16}+5}=\log {}_5\sqrt{25}=1$$

Commented by JDamian last updated on 24/Feb/21

$$\log {}_5\sqrt{25}\ne 2$$$$\log {}_5\sqrt{25}=\log {}_{5}5=1$$

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