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Question Number 62777 by James Bryan Botshabelo last updated on 25/Jun/19

5^(3x−1) .4^(2x−2) =625

$$\mathrm{5}^{\mathrm{3}{x}−\mathrm{1}} .\mathrm{4}^{\mathrm{2}{x}−\mathrm{2}} =\mathrm{625} \\ $$

Commented by kaivan.ahmadi last updated on 25/Jun/19

5^(2x−1) .5^x .4^(2x−1) .4^(−1) =625⇒  20^(2x−1) .5^x =2500  20^x .20^x .20^(−1) .5^x =2500⇒  2000^x =50000⇒  xlog2000=log50000⇒  x=((log50000)/(log2000))=((4+log5)/(3+log2))=((5−log2)/(3+log2))=  =−((−5+log2)/(3+log2))=−((−8+3+log2)/(3+log2))=  (8/(3+log2))−1

$$\mathrm{5}^{\mathrm{2}{x}−\mathrm{1}} .\mathrm{5}^{{x}} .\mathrm{4}^{\mathrm{2}{x}−\mathrm{1}} .\mathrm{4}^{−\mathrm{1}} =\mathrm{625}\Rightarrow \\ $$$$\mathrm{20}^{\mathrm{2}{x}−\mathrm{1}} .\mathrm{5}^{{x}} =\mathrm{2500} \\ $$$$\mathrm{20}^{{x}} .\mathrm{20}^{{x}} .\mathrm{20}^{−\mathrm{1}} .\mathrm{5}^{{x}} =\mathrm{2500}\Rightarrow \\ $$$$\mathrm{2000}^{{x}} =\mathrm{50000}\Rightarrow \\ $$$${xlog}\mathrm{2000}={log}\mathrm{50000}\Rightarrow \\ $$$${x}=\frac{{log}\mathrm{50000}}{{log}\mathrm{2000}}=\frac{\mathrm{4}+{log}\mathrm{5}}{\mathrm{3}+{log}\mathrm{2}}=\frac{\mathrm{5}−{log}\mathrm{2}}{\mathrm{3}+{log}\mathrm{2}}= \\ $$$$=−\frac{−\mathrm{5}+{log}\mathrm{2}}{\mathrm{3}+{log}\mathrm{2}}=−\frac{−\mathrm{8}+\mathrm{3}+{log}\mathrm{2}}{\mathrm{3}+{log}\mathrm{2}}= \\ $$$$\frac{\mathrm{8}}{\mathrm{3}+{log}\mathrm{2}}−\mathrm{1} \\ $$

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