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negative exponents
- Thread starter melcat
- Start date
melcat said:I think I got it, but not sure.
If I raise both sides to -2 then it cancels out the-2 on the x and becomes x= 4^-2 which is 1/4^2 which = 1/16. Is this right?
No. \(\displaystyle (x^{-2})^{-2}\) isn't \(\displaystyle x\), but \(\displaystyle x^4\), because \(\displaystyle (x^a)^b=x^{ab}\).
But you can use your approach - you just need a different power. You need to find some \(\displaystyle b\) that makes \(\displaystyle (x^{-2})^b=x\), that is, \(\displaystyle x^{-2b}=x^1\).
What value of \(\displaystyle b\) makes \(\displaystyle -2b=1\) ?
Then, you can raise both sides to the power of \(\displaystyle b\), and the \(\displaystyle b\) really will cancel out the \(\displaystyle -2\).
It's probably easier, though, to do something like this : (I've changed the numbers)
Q : Solve \(\displaystyle 8y^{-3}=27\).
A : First, I'll move the 8 to the other side, to get \(\displaystyle y^{-3}=\frac{27}{8}\)
Next, I'll remember that \(\displaystyle y^{-a}=\frac{1}{y^a}\), so \(\displaystyle \frac{1}{y^3}=\frac{27}{8}\)
Next, I'll invert both sides of the equation, to get \(\displaystyle y^3=\frac{8}{27}\).
Next, I'll take the cube root of both sides : \(\displaystyle y=\sqrt[3]{\frac{8}{27}}=\frac{\sqrt[3]{8}}{\sqrt[3]{27}}\)
Finally, I'll remember that \(\displaystyle \sqrt[3]{8}=2\) and \(\displaystyle \sqrt[3]{27}=3\), so I get \(\displaystyle y=\frac{2}{3}\)
melcat said:Hi, I have been working on a homework problem for hours and can't figure it out. The problem is x^-2 =4. I know you have to make it 1/x^2 =4, but I don't know where to go from there. Please help. Thank you!
\(\displaystyle \text{Ola . . . There are different ways to solve such problems.}\)
\(\displaystyle \frac{1}{x^2} =4\)
\(\displaystyle 4(x^2) =1\)
\(\displaystyle x^2 =\frac{1}{4}\)
\(\displaystyle x =\pm\frac{1}{2}\)
Hello, melcat!
\(\displaystyle \text{We have: }\:\frac{1}{x^2} \:=\:4\)
\(\displaystyle \text{Multiply by }x^2\!:\quad x^2\left(\frac{1}{x^2}\right) \:=\:x^2(4) \quad\Rightarrow\quad 1 \;=\;4x^2\)
\(\displaystyle \text{Solve for }x\!:\quad 4x^2 \:=\:1 \quad\Rightarrow\quad x^2 \:=\:\tfrac{1}{4} \quad\Rightarrow\quad x \:=\:\pm\sqrt{\tfrac{1}{4}} \quad\Rightarrow\quad\boxed{ x\;=\;\pm\tfrac{1}{2}}\)
\(\displaystyle \text{The problem is: }\:x^{-2} \:=\:4\)
\(\displaystyle \text{I know you have to make it: }\:\frac{1}{x^2}\:=\:4 \quad \text{but I don't know where to go from there.}\)
\(\displaystyle \text{We have: }\:\frac{1}{x^2} \:=\:4\)
\(\displaystyle \text{Multiply by }x^2\!:\quad x^2\left(\frac{1}{x^2}\right) \:=\:x^2(4) \quad\Rightarrow\quad 1 \;=\;4x^2\)
\(\displaystyle \text{Solve for }x\!:\quad 4x^2 \:=\:1 \quad\Rightarrow\quad x^2 \:=\:\tfrac{1}{4} \quad\Rightarrow\quad x \:=\:\pm\sqrt{\tfrac{1}{4}} \quad\Rightarrow\quad\boxed{ x\;=\;\pm\tfrac{1}{2}}\)