Question 159168
Assume the problem is:
{{{8^(2x)*4^(2x-1)}}} = 16
{{{log(8^(2x)) + log(4^(2x-1))}}} = log(16)
log equiv of exponents
{{{2x*log(8) + (2x-1)*log(4)}}} = log(16)
:
Find the logs
.9031(2x) + .6021(2x-1) = 1.2041
:
1.8062x + 1.2042x - .6021 = 1.2041
:
3.01x = 1.2041 + .6021
:
3.01x = 1.8062
x = {{{1.8062/3.01}}}
x = .6
:
:
Check solution on a calc; (8^1.2) * (4^.2) = 16