Question 268217
{{{3^x = 243}}}
So we are trying to find the power of 3 that is equal to 243. Well
{{{3^0 = 1}}}
{{{3^1 = 3}}}
{{{3^2 = 9}}}
{{{3^3 = 27}}}
{{{3^4 = 81}}}
{{{3^5 = 243}}}
So the answer is 5.<br>
If the right side had been 242 or if we din't even think to see if 243 is a whole number power of 3, then we would use logarithms to solve an equation like this. Use a logarithm your calculator "knows", like base 10 or base e (ln):
{{{log((3^x)) = log((243))}}}
Now we use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to move the coefficient out in front. (This property is the reason for using logarithms. It gives us a way to get the variable out of the exponent.):
{{{x*log((3)) = log((243))}}}
Divide both sides by {{{log((3))}}}:
{{{x = log((243))/log((3))}}}
If you use your calculator to find both of these logarithms and then divide them, you will find that the answer is 5 (or a decimal number very, very close to 5. Remember not even your calculator know what log(243) or log(3) are <b>exactly</b>. It must use decimal approximations for them. And these decimal approximations will have round-off errors which may result in an answer like 5.00000001 or 4.9999999.)