Question 149219
Solve for x:
{{{2^((x+1)) * 8^((-x)) = 4}}} Rewrite this as:
{{{2^((x+1))/8^x = 4}}} Multiply both sides by {{{8^x}}}
{{{2^((x+1)) = 4*8^x}}} Recall that:{{{a^n*a^m = a^((m+n))}}}, so you can write:
{{{2^x*2^1 = 4*8^x}}} Divde both sides by 2 to simplify a bit.
{{{2x = 2*8^x}}} ,,,but {{{8^x = (2^3)^x}}}={{{2^((3x))}}} Making this substitution, you get:
{{{2^x = 2*2^((3x))}}} = {{{2^((3x+1))}}} Now, because the bases (2) are equal, the exponents are equal, so...
{{{x = 3x+1}}} Solve for x:
{{{2x = -1}}}
{{{x = -(1/2)}}}
Let's check this solition:
{{{2^((x+1)) * 8^(-x) = 4}}} Substitute {{{x = -(1/2)}}}
{{{2^(((-1/2)+1)) * 8^((1/2)) = 4}}} Simplify the left side.
{{{2^((1/2)) * 8^((1/2)) = 4}}} Substitute{{{8 = 2^3}}}
{{{2^((1/2)) * (2^3)^(1/2) = 4}}} Simplify.
{{{2^((1/2)) * 2^((3)/2) = 4}}} Simplify.
{{{2^(1/2 + 3/2) = 4}}}
{{{2^(4/2) = 4}}}
{{{2^2 = 4}}}
{{{4 = 4}}}