Question 838174
First of all, please put exponents which are not just a variable or single number in parentheses. If you have not also explained your expression in English I would not have been sure what your equation was.<br>
{{{16^(2-n) = (1/4)^(n+1)}}}
Equations with variables in the exponents can usually be solved using logarithms. But if it is possible to rewrite the equation so that each side is a power of the same base, then there is an easier and more accurate way than using logarithms. 16 and 1/4 are powers of each other but what powers these are may not be obvious. But they are both powers of 2 (which should be more obvious). So the solution will be easier if we start by rewriting each side as powers of 2:
{{{(2^4)^(2-n) = (2^(-2))^(n+1)}}}
For powers of a power the rule is multiply the exponents:
{{{2^(8-4n) = 2^(-2n-2)}}}<br>
The next step is based on simple logic. The only way two powers of 2 can be equal, as the now equation says they are, is if the exponents are equal, too. So:
8-4n = -2n-2
Now we solve this simple equation. Adding 4n to each side:
8 = 2n-2
Adding 2:
10 = 2n
Dividing by 2:
5 = n<br>
P.S. If we had used logarithms we might not have gotten an answer of 5. Because of the rounded decimals which a calculator uses, we might have gotten something like 5.00001 or 4.99998. So the method we used above should be used whenever possible.