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.
Write both sides as powers of 1/<em>x</em> :
Recall that 
, so that
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)
14x^2+35x
i think thats how u do it
:)
Let EB and CF be two heights. Then quadrilateral EBCF is rectangle, so BC=EF=5 m. It is given that bases are BC=5 m and AD=11 m. Triangles ABE and DCF are congruent, thus,
Consider right triangle ABE, by the Pythagorean theorem
Answer: 
So first you need to add 4 to each side to get (x+5)^2=16. Next you need to take the square root of each side to get √(x+5)^2=+-√16. After that simplify to get x+5=+-4. Now subtract 5from both sides to look like x=-5+-4. Now just do x=-5+4 and x=-5-4 then solve to get x=-1 and x=-9.
Answer:
1. 81 2. -20 3. 168 4. -24
Step-by-step explanation:
I hope this is right :)
