How do you solve: 3^(x/3) = 5^(x+3) using logs?

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Answer 1 · 2016-04-30T18:01:52

x=-3.88

If $3^{\frac{x}{3}} = 5^{(x + 3)}$ $3^(x/3)=5^((x+3))$

if we take logs (I've used natural logs)
$(\frac{x}{3}) \cdot {log}_{e} 3 = (x + 3) \cdot {log}_{e} 5$ $(x/3)*log_e3=(x+3)*log_e5$

$x \cdot {log}_{e} 3 = (3 x + 9) \cdot {log}_{e} 5$ $x*log_e3=(3x+9)*log_e5$

$x \cdot {log}_{e} 3 - 3 x \cdot {log}_{e} 5 = 9 \cdot {log}_{e} 5$ $x*log_e3-3x*log_e5=9*log_e5$

$x \cdot ({log}_{e} 3 - 3 \cdot {log}_{e} 5) = 9 \cdot {log}_{e} 5$ $x*(log_e3-3*log_e5)=9*log_e5$

$x \cdot ({log}_{e} 3 - {log}_{e} 5^{3}) = 9 \cdot {log}_{e} 5$ $x*(log_e3-log_e5^3)=9*log_e5$

$x \cdot ({log}_{e} 3 - {log}_{e} 125) = 9 \cdot {log}_{e} 5$ $x*(log_e3-log_e125)=9*log_e5$

$x \cdot {log}_{e} (\frac{3}{125}) = 9 \cdot {log}_{e} 5$ $x*log_e(3/125)=9*log_e5$

$x = 9 \cdot \frac{{log}_{e} 5}{{log}_{e} (\frac{3}{125})} = - 3.88$ $x=9*log_e5/(log_e(3/125))=-3.88$