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Question Number 149410 by mathdanisur last updated on 05/Aug/21

$$\frac{{log}_2{2}^{20}+{log}_{2}20\cdot {log}_{2}5-2{log}_2{2}^5}{{log}_{2}20+2{log}_{2}5}=?$$

Answered by Rasheed.Sindhi last updated on 05/Aug/21

$$\frac{{log}_2{2}^{20}+{log}_{2}20\cdot {log}_{2}5-2{log}_2{2}^5}{{log}_{2}20+2{log}_{2}5}$$$$=\frac{20+{log}_{2}20\cdot {log}_{2}5-2\left(5\right)}{{log}_{2}20+2{log}_{2}5}$$$$=\frac{10+{log}_2(2^2.5)\cdot {log}_{2}5}{{log}_2(2^2.5)+2{log}_{2}5}$$$$=\frac{10+({log}_2{2}^2+{log}_{2}5)\cdot {log}_{2}5}{{log}_2{2}^2+{log}_{2}5+2{log}_{2}5}$$$$=\frac{10+(2+{log}_{2}5)\cdot {log}_{2}5}{2+3{log}_{2}5}$$$$=\frac{10+2{log}_{2}5+{log}_{2}5\cdot {log}_{2}5}{2+3{log}_{2}5}$$$$=\frac{10+2{log}_{2}5+{\left({log}_{2}5\right)}^2}{2+3{log}_{2}5}$$$$=\frac{10+2(2.3219)+{(2.3219)}^2}{2+3(2.3219)}\approx 2.2346$$

Commented by mathdanisur last updated on 05/Aug/21

$$ThankYou\mathrm{Ser}$$

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