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Alex777 [14]
3 years ago
12

Solve for x using logs- 2*7^x=17*3^x

Mathematics
1 answer:
horsena [70]3 years ago
8 0

2\cdot7^x=17\cdot3^x\ \ \ \ |\log\\\\\log(2\cdot7^x)=\log(17\cdot3^x)\ \ \ \ |use\ \log (x\cdot y)=\log x+\log y\\\\\log2+\log7^x=\log17+\log3^x\ \ \ \ |-\log2;\ |-\log3^x\\\\\log7^x-\log3^x=\log17-\log2\ \ \ \ |use\ \loga^n=n\log a\\\\x\log7-x\log3=\log17-\log2\\\\x(\log7-\log3)=\log17-\log2\ \ \ |use\ \log x-\log y=\log\dfrac{x}{y}\\\\x\log\dfrac{7}{3}=\log\dfrac{17}{2}\ \ \ \ |:\log\dfrac{7}{3}\\\\\boxed{x=\dfrac{\log\frac{17}{2}}{\log\frac{7}{3}}}

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