Question 377487
{{{3(9)^(x-1)=(81)^(2x+1)}}}
Since the 3, the 9 and the 81 are all powers of 3, we can rewrite each side of this equation as powers of 3. This is probably the easiest way to solve this equation.
{{{3(3^2)^(x-1)=(3^4)^(2x+1)}}}
On each side we have a power to a power. The exponent rule for this is to multiply the exponents. This gives us:
{{{3(3^(2x-2))=3^(8x+4)}}}
On the left side we have 3 (or {{{3^1}}}) times a power of three. The exponent rule for this is to add the exponents:
{{{3^(1+2x-2)=3^(8x+4)}}}
which simplifies to:
{{{3^(2x-1)=3^(8x+4)}}}
We now have both sides as powers of 3. The only way these powers of 3 can be equal if if the exponents are equal. So:
2x-1 = 8x+4
This is an easy equation to solve. Subtract 2x from each side:
-1 = 6x + 4
Subtract 4 from each side:
-5 = 6x
Divide both sides by 6:
{{{-5/6 = x}}}
And we're done.<br>
This problem can also be done with logarithms. And if the 3, the 9 and the 81 had not all been powers of the same number we would have to use logarithms. We could use base 3 logarithms:
{{{log(3, (3(9)^(x-1)))=log(3, (81)^(2x+1))}}}
{{{log(3, (3)) + log(3, ((9)^(x-1)))=log(3, (81^(2x+1)))}}}
{{{log(3, (3)) + (x-1)log(3, (9))= (2x+1)log(3, (81))}}}
{{{log(3, (3)) + x*log(3, (9)) - log(3, (9)) = 2x*log(3, (81)) + log(3, (81))}}}
{{{log(3, (3)) - log(3, (9)) - log(3, (81)) = 2x*log(3, (81)) - x*log(3, (9))}}}
{{{log(3, (3)) - log(3, (9)) - log(3, (81)) = x*(2*log(3, (81)) - log(3, (9)))}}}
{{{(log(3, (3)) - log(3, (9)) - log(3, (81)))/(2*log(3, (81)) - log(3, (9))) = x}}}
Since {{{3 = 3^1}}}, {{{9 = 3^2}}} and {{{81 = 3^4}}} the base 3 logs of each are 1, 2 and 4, respectively. Substituting these in to the equation we get:
{{{(1 - 2 - 4)/(2*4 - 2) = x}}}
{{{(-5)/6 = x}}}<br>
We could also use base 10 logarithms:
{{{log((3(9)^(x-1)))=log((81^(2x+1)))}}}
{{{log((3)) + log(((9)^(x-1)))=log((81^(2x+1)))}}}
{{{log((3)) + (x-1)log((9))= (2x+1)log((81))}}}
{{{log((3)) + x*log((9)) - log((9)) = 2x*log((81)) + log((81))}}}
{{{log((3)) - log((9)) - log((81)) = 2x*log((81)) - x*log((9))}}}
{{{log((3)) - log((9)) - log((81)) = x*(2*log((81)) - log((9)))}}}
{{{(log((3)) - log((9)) - log((81)))/(2*log((81)) - log((9))) = x}}}
We could now get out our calculators and find a decimal approximation for x. This should work out to something very close to the decimal form of -5/6: -0.8333333....