How do you solve ${(\frac{3}{4})}^{2 x} = \frac{64}{27}$ $(3/4)^(2x) = 64/27$ ?

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How do you solve ${(\frac{3}{4})}^{2 x} = \frac{64}{27}$ $(3/4)^(2x) = 64/27$ ?

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Answer 1 · 2016-03-28T11:47:57

$x = - 5, 3188416645$ $x=-5,3188416645$

Explanation:

${(\frac{3}{4})}^{2 x} = \frac{64}{27}$ $(3/4)^(2x)=64/27$
${(\frac{3}{4})}^{2 x} = \frac{2^{6}}{3^{3}}$ $(3/4)^(2x)=2^6/3^3$

$\frac{3^{2 x}}{4^{2 x}} = \frac{2^{6}}{3^{3}}$ $3^(2x)/4^(2x)=2^6/3^3$

$3^{2 x} \cdot 3 = 4^{2 x} \cdot 2^{6}$ $3^(2x)*3=4^(2x)*2^6$

$3^{2 x + 1} = 2^{4 x} \cdot 2^{6}$ $3^(2x+1)=2^(4x)*2^6$

$3^{2 x + 1} = 2^{4 x + 6}$ $3^(2x+1)=2^(4x+6)$

${log 3}^{2 x + 1} = {log 2}^{4 x + 6}$ $log 3^(2x+1)=log 2^(4x+6)$

$(2 x + 1) \cdot log 3 = (4 x + 6) \cdot log 2$ $(2x+1)*log 3=(4x+6)*log 2$
$log 3 = 0, 47712125773$ $log 3=0,47712125773$
$log 2 = 3010299957$ $log 2=3010299957$
$(2 x + 1) \cdot 0, 47712125773 = (4 x + 6) \cdot 0, 3010299957$ $(2x+1)*0,47712125773=(4x+6)*0,3010299957$
$0, 9542425094 x + 0, 47712125773 = 1, 2041199827 x + 1, 806179974$ $0,9542425094x+0,47712125773=1,2041199827x+1,806179974$
$0, 2498774733 x = - 1, 329058716$ $0,2498774733x=-1,329058716$

$x = - 5, 3188416645$ $x=-5,3188416645$

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How do you solve ${(\frac{3}{4})}^{2 x} = \frac{64}{27}$ $(3/4)^(2x) = 64/27$ ?

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How do you solve ${(\frac{3}{4})}^{2 x} = \frac{64}{27}$ $(3/4)^(2x) = 64/27$ ?