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Question Number 56418 by gunawan last updated on 16/Mar/19

$$\mathrm{The}\mathrm{value}\mathrm{of}{a}^{{\log }_{b}x}\mathrm{where}a=0.2,b=\sqrt{5,}$$$$x=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...\mathrm{to}\mathrm{\infty }\mathrm{is}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19

$$x=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{1}{2}=0.5$$$$t={log}_{b}x\to b^t=x\to {\left(5\right)}^{\frac{t}{2}}=\frac{1}{2}=0.5$$$$\frac{t}{2}={log}_5(0.5)\to t=2{log}_5(0.5)$$$$so{a}^{{log}_{b}x}\to 0.2^{{log}_5(0.25)}$$$$$$$$$$$$$$$$$$

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