Question #6daa9

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Question #6daa9

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Answer 1 · 2016-04-05T04:38:19

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

Explanation:

Take the natural logarithm on both sides

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

From the identity

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

We can simplify the above equation as

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

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

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

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

Answer 2 · 2016-04-05T08:48:32

A slightly different approach

$\Rightarrow x = \frac{ln (2) + 2}{5}$ $=>x = (ln(2)+2)/5$

Explanation:

Given: $e^{3 x} \times e^{2 x - 1} = 2 e$ $" "e^(3x) xx e^(2x-1)=2e$

Compare to $10^{2} \times 10^{1} = 10^{3} = 10^{2 + 1}$ $10^2xx10^1 = 10^3 =10^(2+1)$

Using the above method we have;

$e^{3 x} \times e^{2 x - 1} = 2 e \to e^{3 x + 2 x - 1} = 2 e$ $" "e^(3x) xx e^(2x-1)=2e" "->" "e^(3x+2x-1)=2e$

Divide both sides by $e$ $e$

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

$\Rightarrow e^{5 x - 2} = 2$ $=>e^(5x-2)=2$

Take logs of both sides

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

But $ln (e) = 1$ $ln(e)=1$

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

$\Rightarrow x = \frac{ln (2) + 2}{5}$ $=>x=(ln(2)+2)/5$

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Question #6daa9

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