Question

What value of n solves the equation?

http://static.k12.com/bank_packages/files/media/mathml_5340dc3aef80a9557850126719afe511604c87bc_1.gif

n = ______

Answer 1

Anyone

Answer 2

This is called an "exponential equation", because we can see an equal sign and the variable is on the exponent, so we will recurr to the exponential properties and maybe some logarithmic properties as qell in order to solve this, I will suppose you know these properties since they are the ground for a lot of algebraic and mathematical properties.

The strategy is to use the propeties of exponents and logarithms and make the variable appear in the numerator (in other words, to move it away from the exponent).
So, let's begin:
\[3^n = \frac{ 1 }{ 81 }\]
So, let's use the property of the inverse, that'll just mean I can take the "81" to the numerator and make it's exponent negaive:
\[3^n=81^{-1}\]
This is where much people don't know what to do, usually the teachers say "convert the 81 into a base whose exponent is x", true, but we will take the logarithm with base "3" on both side of the equality, to make it simpler:
\[\log_{3}3^n =\log_{3}81^{-1} \]
So therefore:
\[nlog_{3}3 = -\log_{3}81 \]
but Logarithm with base 3 of 3 is 1, so therefore:
\[n=-\log_{3}81 \]
That is a solution, I'll leave to you the duty of determining what Logarithm with base 3 of 81 is worth.