How do you simplify and solve $(4^{2 x} + 3) + (4^{2 x} + 2) = 320$ $(4^(2x)+3) + (4^(2x)+2) = 320$ ?

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How do you simplify and solve $(4^{2 x} + 3) + (4^{2 x} + 2) = 320$ $(4^(2x)+3) + (4^(2x)+2) = 320$ ?

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Answer 1 · 2015-04-04T21:47:01

$4^{2 x} \cdot 2 = 315$ $4^(2x) *2=315$

$\to 4^{2 x} = \frac{315}{2}$ $-> 4^(2x) = 315/2$

$\to ln (4^{2 x}) = ln (\frac{315}{2})$ $->ln(4^(2x)) = ln(315/2)$

$\to 2 x ln (4) = ln (157.5)$ $->2xln(4)=ln(157.5)$

$\to 2 x = \frac{ln (157.5)}{ln (4)}$ $->2x=ln(157.5)/ln(4)$

$\to 2 x ≅ 3.64$ $-> 2x ~=3.64$

$\to x ≅ 1.82$ $-> x~=1.82$

Answer 2 · 2015-04-04T21:55:23

Have a look:
enter image source here
At the end I used the change of base of logs where you have:
${log}_{a} (c) = \frac{ln (c)}{ln (a)}$ $log_a(c)=ln(c)/ln(a)$

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How do you simplify and solve $(4^{2 x} + 3) + (4^{2 x} + 2) = 320$ $(4^(2x)+3) + (4^(2x)+2) = 320$ ?

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