<PRE><B>Question 723725</B>
Rewrite each as an equivalent exponential equation.
1) {{{log(3,5) = y}}}
The base of 3, raised to the power of y, equals 5
{{{3^y = 5}}}
:
2 {{{log(a,7) =-2}}}
{{{a^-2 = 7}}}
:
Solve
{{{log(2,x) = -3}}}
{{{2^-3 = x}}}
The reciprocal gets rid of the neg exponent
{{{1/(2^3) = x}}}
{{{1/8 = x}}}
:
{{{log(x,9) = 1/2}}}
{{{x^(1/2) = 9 }}}
square both sides
{{{x^((1/2)*2) = 9^2}}}
{{{x^1 = 81}}}
or just
x = 81</PRE>