View Full Version : Question on exponent laws
1. The problem statement, all variables and given/known data
2^3 / 2^0 -2^-1
2. Relevant equations
solving using exponent laws
3. The attempt at a solution
i move the negetive one over and make it 2^3 /2^0 = 1/2^1
I get 8 on the left side, and 2 on the right, bring the 6 from the right to the left and get 2. The answer is 16, i dont know how to get it. Anyone?
QuarkCharmer
Jul7-11, 08:03 PM
2^3 / 2^0 -2^-1
Is that:
\frac{2^3}{2^{0}-2^{-1}} ?
fluidistic
Jul7-11, 08:06 PM
2^3 / 2^0 -2^-1
Is that:
\frac{2^3}{2^{0}-2^{-1}} ?
Yes because the answer is 16 :)
HallsofIvy
Jul7-11, 08:10 PM
\frac{2^3}{2^0}- 2^{-1}= \frac{8}{1}- \frac{1}{2}= 8- \frac{1}{2}= \frac{15}{2}
\frac{2^3}{2^0- 2^{-1}}= \frac{8}{1-\frac{1}{2}}= \frac{8}{\frac{1}{2}}= 16
How do you solve this problem?
3^g+3 - 3^g+2 = 1458
Ive been stuck on it for a whole hour
Ray Vickson
Jul7-11, 08:51 PM
How do you solve this problem?
3^g+3 - 3^g+2 = 1458
Ive been stuck on it for a whole hour
There is no solution. You write 3^g+3 - 3^g+2, which means (3^g) + 3 - (3^g) + 2, and this is equal to 5. If you actually mean something different, USE BRACKETS. For example, maybe you mean 3^(g+3) - 3^(g+2) = 1458. Of course, that would have a solution, which we could get quite easily using the fact that 3^(g+3) = 3*3^(g+2).
RGV
Theres no brackets, and there is a solution , its 4.
Heres another one.
-500 = 5^y+1 -5^y+2
looks like the same type of problem, answer to this ones 3
Ray Vickson
Jul7-11, 11:00 PM
You seem to be missing a VERY IMPORTANT point: you *must* use brackets because otherwise what you are writing is in violation of all standard math writing rules. People will automatically assume you mean what you write, but apparently that is not true!
RGV
Theres no brackets, and there is a solution , its 4.
Here's another one.
-500 = 5^y+1 - 5^y+2
looks like the same type of problem, answer to this ones 3
If you mean that y+1 and y+2 are exponents, then you should (must, actually) use parentheses or other grouping symbols to indicate where the exponent ends and the rest of the expression picks up again.
I assume you meant:
-500 = 5^(y+1) - 5^(y+2) .
It works even better to use the X2 button above the advanced version of the reply box. Then you don't need the parentheses. This gives:
-500 = 5y+1 - 5y+2
To solve this, notice that 5y+2 = (5)(5y+1)
Factor (5y+1) out of 5y+1 - (5)(5y+1).
The answer is y = 2.
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