Question 423081
{{{y= sqrt(9^(x-1))}}}
Since the desired form has no square roots, we will start by eliminating the square root. We can replace the square root with an exponent of 1/2. Because of occasional display problems with fractional exponents I am going to use the decimal equivalent of 1/2, 0.5:
{{{y= (9^(x-1))^0.5}}}
We now have a power of a power of 8. The rule for exponents for this is to multiply the exponents:
{{{y=  9^((x-1)*0.5)}}}
We still want an exponent of just x. So we need still to eliminate the -1 and the 0.5. Since {{{9 = 3^2}}} we can introduce a 2 into the exponent:
{{{y=  (3^2)^((x-1)*0.5)}}}
Again we use the rule for exponents:
{{{y=  3^(2*(x-1)*0.5)}}}
The exponent simplifies to:
{{{y=  3^(x-1)}}}
We're making progress. All we have to do now is eliminate the -1. The exponent is now a subtraction. And when do we subtract exponents? Answer: When we divide. So if we "undo" a division involving {{{3^x}}} and {{{3^1}}}...
{{{y=  3^x/3^1}}}
The denominator simplifies:
{{{y=  3^x/3}}}
The exponent is finally what we want it to be. But we want something times {{{3^x}}} not {{{3^x}}} divided by something. Since division by 3 is the same as multiplying by the reciprocal of 3 this is an easy checge:
{{{y=  (1/3)3^x}}}
We finally have the desired form with the "a" being 1/3 and the "b" being 3.<br>
P.S. In response to the question in your "Thank you"...
The problem asked for an equation of the form:
{{{y = a*b^x}}}
With the steps I've shown above we've transformed
{{{y= sqrt(9^(x-1))}}}
into
{{{y = (1/3)*3^x}}}
which is the desired form. So I don't understand your question: "Does x stay the exponent?" First of all, the desired form wants the exponent to be x so we don't want the exponent to stop being x. Second, I'm not sure where the "x" can go. (Remember, the order of operations (aka PEMDAS) requires that we raise 3 to the x power before we multiply by 1/3. So the 1/3 and the 3 do not cancel.)