For the logarithms: $2^{x + 1} = 3^{x}$ $2^(x+1) = 3^x$ , how do you solve for x?

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Answer 1 · 2015-10-04T03:37:12

$x = \frac{ln (2)}{ln (3) - ln (2)}$ $x = ln(2)/(ln(3) - ln(2))$

Take the log of the two sides

$ln (2^{x + 1}) = ln (3^{x})$ $ln(2^(x+1)) = ln(3^x)$

Pass the exponents to the front of the log

$(x + 1) ln (2) = x ln (3)$ $(x+1)ln(2) = xln(3)$

Expand the left side

$x ln (2) + ln (2) = x ln (3)$ $xln(2) + ln(2) = xln(3)$

Isolate x

$ln (2) = x ln (3) - x ln (2)$ $ln(2) = xln(3)-xln(2)$

Put x in evidence

$ln (2) = x (ln (3) - ln (2))$ $ln(2) = x(ln(3)-ln(2))$

Pass that dividing

$x = \frac{ln (2)}{ln (3) - ln (2)}$ $x = ln(2)/(ln(3) - ln(2))$

Or, if you want, you can use log properties to change that into other logs, like

$x = \frac{ln (2)}{ln (\frac{3}{2})}$ $x = ln(2)/(ln(3/2))$ or $x = {log}_{\frac{3}{2}} (2)$ $x = log_(3/2)(2)$

For the logarithms: 2x+1=3x2^(x+1) = 3^x, how do you solve for x?