Question 718243
First of all, if you put multiple-term exponents in parentheses:
5^(6x+3)=37
you don't have to explain what the exponent is.<br>
{{{5^(6x+3)=37}}}
To find a decimal approximation for the solution we will be using logarithms, specifically logs your calculator "knows" (base e, "ln", or base 10, "log):
{{{ln(5^(6x+3))=ln(37)}}}
Next we use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, which allows us to move the exponent of the argument out in front (where can then "get at" the variable):
{{{(6x+3)*ln(5)=ln(37)}}}
Dividing by ln(5):
{{{6x+3=ln(37)/ln(5)}}}
Subtract 3:
{{{6x=ln(37)/ln(5)-3}}}
Divide by 6:
{{{x=(ln(37)/ln(5)-3)/6}}}
This is an exact expression for the solution to your equation. For a decimal approximation, use your calculator. (Note: {{{ln(37)/ln(5)}}} is <u>not</u> the same as {{{ln(37/5)}}}!! So you have to find the two ln's separately and then divide, then subtract 3 and finally divide by 6.)