Question 135430
Solve for x:
{{{log(5,(5-4x)) = log(sqrt(5),(2-x))}}}
Change the base of the second logarithm to that of the first using:{{{log(b,(M)) = (log(a,(M)))/(log(a,(b)))}}}
{{{log(sqrt(5),(2-x)) = (log(5,(2-x)))/(log(5,(sqrt(5))))}}}={{{(log(5,(2-x)))/(1/2)log(5,(5))}}}={{{2*log(5,(2-x))}}}
So, we end up with:
{{{log(5,(5-4x)) = 2*log(5,(2-x))}}} Applying the power rule to the right side:
{{{log(5,(5-4x)) = log(5,(2-x)^2)}}} therefore:
{{{5-4x = (2-x)^2}}} Simplifying:
{{{5-4x = 4-4x+x^2}}} Add 4x to both sides.
{{{5 = 4+x^2}}} Subtract 5 from both sides.
{{{0 = x^2-1}}} Factor the right side.
{{{(x+1)(x-1) = 0}}} so...
{{{x = 1}}} or {{{x = -1}}}
The answer can be read as: x = + or - 1