How do you solve $5^{x - 1} = 3^{x}$ $5^(x - 1) = 3^x$ ?

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Answer 1 · 2016-02-11T04:09:41

You must first convert to logarithmic form.

${log 5}^{x - 1} = {log 3}^{x}$ $log5^(x - 1) = log3^x$

$(x - 1) log 5 = x log 3$ $(x - 1)log5 = xlog3$

Distribute the parentheses:

$x log 5 - log 5 = x log 3$ $xlog5 - log5 = xlog3$

Put all x terms to the left side of the equation:

$x log 5 - x log 3 = log 5$ $xlog5 - xlog3 = log5$

Factor out the x:

$x (log 5 - log 3) = log 5$ $x(log5 - log3) = log5$

Use the quotient rule:

$x (log (\frac{5}{3})) = log 5$ $x(log(5/3)) = log5$

$x = \frac{log 5}{log (\frac{5}{3})}$ $x = log5/(log(5/3)$

$x = {log}_{\frac{5}{3}} 5$ $x = log_(5/3)5$

Practice exercises:

a) $2^{x - 2} = 3^{2 x + 4}$ $2^(x - 2) = 3^(2x + 4)$

b) $3^{3 x} = 4 \times 5^{x - 6}$ $3^(3x) = 4 xx 5^(x - 6)$

Good luck!

How do you solve 5^(x - 1) = 3^x5x−1=3x 5^(x - 1) = 3^x?