How do you solve $8^{x} = 4 \times 12^{2 x}$ $8^x = 4 times 12^(2x)$ ?

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Answer 1 · 2016-09-16T12:17:18

$x = - \frac{2 ln (2)}{ln (18)}$ $x = - (2 ln(2)) / (ln(18))$

We have: $8^{x} = 4 \times 12^{2 x}$ $8^(x) = 4 times 12^(2 x)$

Let's apply the natural logarithm to both sides of the equation:

$\Rightarrow ln (8^{x}) = ln (4 \times 12^{2 x})$ $=> ln(8^(x)) = ln(4 times 12^(2 x))$

Using the laws of logarithms:

$\Rightarrow x ln (8) = ln (4) + 2 x ln (12)$ $=> x ln(8) = ln(4) + 2 x ln(12)$

Let's express some numbers in terms of $2$ $2$ :

$\Rightarrow x ln (2^{3}) = ln (2^{2}) + 2 x ln (12)$ $=> x ln(2^(3)) = ln(2^(2)) + 2 x ln(12)$

$\Rightarrow 3 x ln (2) = 2 ln (2) + 2 x ln (12)$ $=> 3 x ln(2) = 2 ln(2) + 2 x ln(12)$

$\Rightarrow 3 x ln (2) - 2 x ln (12) = 2 ln (2)$ $=> 3 x ln(2) - 2 x ln(12) = 2 ln(2)$

$\Rightarrow x (3 ln (2) - 2 ln (12)) = 2 ln (2)$ $=> x (3 ln (2) - 2 ln(12)) = 2 ln(2)$

$\Rightarrow x (ln (2^{3}) - ln (12^{2})) = 2 ln (2)$ $=> x (ln(2^(3)) - ln (12^(2))) = 2 ln(2)$

$\Rightarrow x (ln (\frac{2^{3}}{12^{2}})) = 2 ln (2)$ $=> x (ln((2^(3)) / (12^(2)))) = 2 ln(2)$

$\Rightarrow x (ln (\frac{8}{144})) = 2 ln (2)$ $=> x (ln((8) / (144))) = 2 ln(2)$

$\Rightarrow x (ln (\frac{1}{18})) = 2 ln (2)$ $=> x (ln ((1) / (18))) = 2 ln(2)$

$\Rightarrow x (ln (1) - ln (18)) = 2 ln (2)$ $=> x (ln(1) - ln(18)) = 2 ln(2)$

$\Rightarrow x (0 - ln (18)) = 2 ln (2)$ $=> x (0 - ln (18)) = 2 ln(2)$

$\Rightarrow - x ln (18) = 2 ln (2)$ $=> - x ln(18) = 2 ln(2)$

$\Rightarrow x ln (18) = - 2 ln (2)$ $=> x ln(18) = - 2 ln(2)$

$\Rightarrow x = - \frac{2 ln (2)}{ln (18)}$ $=> x = - (2 ln(2)) / (ln(18))$

Therefore, the solution to the equation is $x = - \frac{2 ln (2)}{ln (18)}$ $x = - (2 ln(2)) / (ln(18))$ .

How do you solve 8^x = 4 times 12^(2x)8x=4×122x8^x = 4 times 12^(2x)?