The base of a logarithm should always be positive and can't be equal to 1, so the domain is 0 < <em>x</em> < 1 or <em>x</em> > 1.
![\log_{\frac1x}243=5](https://tex.z-dn.net/?f=%5Clog_%7B%5Cfrac1x%7D243%3D5)
Write both sides as powers of 1/<em>x</em> :
![\left(\dfrac1x\right)^{\log_{\frac1x}243}=\left(\dfrac1x\right)^5](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac1x%5Cright%29%5E%7B%5Clog_%7B%5Cfrac1x%7D243%7D%3D%5Cleft%28%5Cdfrac1x%5Cright%29%5E5)
Recall that
, so that
![243=\left(\dfrac1x\right)^5](https://tex.z-dn.net/?f=243%3D%5Cleft%28%5Cdfrac1x%5Cright%29%5E5)
![243=\dfrac1{x^5}](https://tex.z-dn.net/?f=243%3D%5Cdfrac1%7Bx%5E5%7D)
![x^5=\dfrac1{243}](https://tex.z-dn.net/?f=x%5E5%3D%5Cdfrac1%7B243%7D)
Take the 5th root of both sides, recalling that 3⁵ = 243, so
![x=\sqrt[5]{\dfrac1{243}}=\boxed{\dfrac13}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B5%5D%7B%5Cdfrac1%7B243%7D%7D%3D%5Cboxed%7B%5Cdfrac13%7D)
The numbers you are thinking of are 161 and 125.
rise over run
(-1,2) (2,2)
2-2/2-(-1)
2-2/2+1 equals 0/3
or alternatively
2-2/-1 - 2 which would be -0/3.
Whichever point is the first helps determine this alot
rise/run=rise over run= y2 - y1/ x2 - x1
Answer: length = 8cm
Width = 4cm
Step-by-step explanation:
Let the width of the rectangle be represented by x.
Since the length of a rectangle is twice as long as the width of the rectangle, then the length will be: = 2 × x = 2x
Note that, Area of a rectangle = Length × Width
Therefore,
2x × x = 32
2x² = 32
Divide both side by 2
2x²/2 = 32/2
x² = 16
x = ✓16
x = 4
Therefore, width = 4cm
Since length = 2x
Length = 2 × 4cm = 8cm
Check: length × width = Area
8cm × 4cm = 32cm²