Find X, im assuming?
\(2^x*5^{3x} = 4^{2x+1}\)
Lets go ahead and take the natural log of each side
\(ln{2^x}*ln{5^{3x}} =ln{ 4^{2x+1}}\)
Then bring down those exponets and pull out the x's on the left
\(x(ln{2}*(3)ln{5}) = (2x+1)*ln{4}\)
Pull the X's over to the right and natural log's to the left
\(\frac{ln{2}*(3)ln{5}}{ln{4}} = \frac{2x+1}{x}\)
Simplify
(ln(2)+3(ln(5)))/ln(4) = 3.98289
(2x+1)/x
You can split these into 2x/x and 1/x
becomes 2+1/x
3.98289 = 2+1/x
1.98289 = 1/x
1.98289x = 1
x = 1/1.98289 = 0.5043
x = .5043
