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Question Number 188218 by pascal889 last updated on 26/Feb/23

Answered by cortano12 last updated on 27/Feb/23

$$\mathrm{let}\log {}_{\mathrm{x}}10=\frac{1}{\log {}_{10}\mathrm{x}}=\mathrm{p},\mathrm{x}>0,\mathrm{x}\ne 1$$$$\Rightarrow 2.\log {}_{\mathrm{x}}10-\frac{1}{2}\left[\log {}_{\mathrm{x}}\left(\frac{\sqrt{\mathrm{x}}}{100}\right)\right]-\frac{\left(\frac{1}{2}\right)}{1-\log {}_{\mathrm{x}}10}=-\frac{7}{2}$$$$\Rightarrow 2\mathrm{p}-\frac{1}{2}(\frac{1}{2}-2\mathrm{p})-\frac{1}{2-2\mathrm{p}}=-\frac{7}{2}$$$$\Rightarrow 4\mathrm{p}-\frac{1}{2}+2\mathrm{p}-\frac{1}{4-4\mathrm{p}}=-7$$$$\Rightarrow 6\mathrm{p}-\frac{1}{4-4\mathrm{p}}=-\frac{13}{2}$$$$\Rightarrow \mathrm{p}=\frac{\sqrt{601}-1}{24}$$$$\Rightarrow \log {}_{\mathrm{x}}\left(10\right)=\frac{\sqrt{601}-1}{24}$$$$\Rightarrow \frac{1}{\log {}_{10}\left(\mathrm{x}\right)}=\frac{\sqrt{601}-1}{24}$$$$\Rightarrow \mathrm{x}={10}^{\frac{24}{\sqrt{601}-1}}$$

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