Question #6daa9

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Answer 1 · 2016-04-05T04:38:19

$x = frac{ln(2) + 2}{5}$

Take the natural logarithm on both sides

$ln(e^{3x} * e^{2x-1}) = ln(2e)$

From the identity

$ln(ab) = ln(a) + ln(b)$

We can simplify the above equation as

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

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

$5x = ln(2) + 2$

$x = frac{ln(2) + 2}{5}$

Answer 2 · 2016-04-05T08:48:32

A slightly different approach

$=>x = (ln(2)+2)/5$

Given: $" "e^(3x) xx e^(2x-1)=2e$

Compare to $10^2xx10^1 = 10^3 =10^(2+1)$

Using the above method we have;

$" "e^(3x) xx e^(2x-1)=2e" "->" "e^(3x+2x-1)=2e$

Divide both sides by $e$

$e^(3x+2x-2)=2$

$=>e^(5x-2)=2$

Take logs of both sides

$(5x-2)ln(e)=ln(2)$

But $ln(e)=1$

$5x-2=ln(2)$

$=>x=(ln(2)+2)/5$