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Question Number 159507 by cortano last updated on 18/Nov/21

Commented by tounghoungko last updated on 18/Nov/21

$$\frac{5.5^{-2x}+5^3.5^{-2x}}{\frac{5}{8}.2^{2x}}=4$$$$\iff 5^{-2x}+5^2.5^{-2x}=\frac{1}{2}.2^{2x}$$$$\iff 26.5^{-2x}=\frac{1}{2}.2^{2x}$$$$\iff 52={10}^{2x}\Rightarrow x=\frac{1}{2}\log {}_{10}\left(52\right)$$

Answered by Tokugami last updated on 18/Nov/21

$$\frac{5^{1-2x}+\sqrt{{25}^{3-2x}}}{2^{2x-1}+2^{2x-3}}=4$$$$\frac{5^{1-2x}+\sqrt{{25}^{3-2x}}}{2^{2x-1}+2^{2x-3}}=2^2$$$$5^{1-2x}+\sqrt{{25}^{3-2x}}=2^{2x-1+2}+2^{2x-3+2}$$$$5^{1-2x}+5^{3-2x}=2^{2x+1}+2^{2x-1}$$$$5^{-2x}(5^1+5^3)=2^{2x}(2^1+2^{-1})$$$$\frac{130}{5^{2x}}=2^{2x}\left(\frac{5}{2}\right)$$$$\frac{260}{5}={10}^{2x}$$$$52={10}^{2x}$$$$\ln \left(52\right)=2x\ln \left(10\right)$$$$2x=\frac{\ln 52}{\ln 10}={\log }_{10}\left(52\right)$$$$x=\frac{1}{2}{\log }_{10}\left(52\right)$$$$x\approx 0.858$$

Answered by qaz last updated on 18/Nov/21

$$\frac{5^{1-2\mathrm{x}}\sqrt{{25}^{3-2\mathrm{x}}}}{2^{2\mathrm{x}-1}+2^{2\mathrm{x}-3}}=\frac{5^{1-2\mathrm{x}}+5^{3-2\mathrm{x}}}{2^{2\mathrm{x}-1}+2^{2\mathrm{x}-3}}=\frac{(5+5^3)5^{-2\mathrm{x}}}{(2^{-1}+2^{-3})2^{2\mathrm{x}}}=208\cdot {10}^{-2\mathrm{x}}=4$$$$\Rightarrow \mathrm{x}=\frac{1}{2}\lg 52$$

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