Re: 8 problems that I can't understand after trying for days and looking examples onl

Quote:

Originally Posted by

**abender** $\displaystyle f(x)=(x-2)^3$

$\displaystyle y=(x-2)^3$

Switching $\displaystyle x$ and $\displaystyle y$:

$\displaystyle x = (y-2)^3$

Now, to solve for $\displaystyle y$, raise both sides to the $\displaystyle \tfrac{1}{3}$ power:

$\displaystyle x^{\frac{1}{3}} = ((y-2)^3)^{\frac{1}{3}}$

$\displaystyle x^{\frac{1}{3}} = y-2 \implies y = x^{\frac{1}{3}}+2$

REMEMBER, the cube root is the same as the one-third power, just like the square root of a number is the same as the number raised to the one-half.

Oh, I get it. The "cube root is the same as one-third power" messed me up. Way clearer now. I was trying to do it with: (y-2)(y-2)(y-2).

Quote:

Originally Posted by

**abender** Do you mean $\displaystyle y^{2}x$?

Yes.

Re: 8 problems that I can't understand after trying for days and looking examples onl

**Question 8:**

A simple Google search for "log properties" yields Logarithmic Properties -- LEARN ALL OF THESE!

**These properties must be understood to solve this problem:**

1) $\displaystyle \log_b(xy) = \log_b(x) + \log_b(y)$

2) $\displaystyle \log_b(\tfrac{x}{y}) = \log_b(x) - \log_b(y)$

3) $\displaystyle \log_b(x^n) = n\log_b(x)$

First, change $\displaystyle \displaystyle^8\sqrt{7}$ to $\displaystyle 7^{\tfrac{1}{8}}$, which I explained earlier.

$\displaystyle \small{\log_{15}\frac{7^{\frac{1}{8}}}{y^2x} = \color{red}\log_{15}(7^{\frac{1}{8}}) \color{black} - \color{blue}\log_{15}(y^2x)\color{black} = \color{red}\frac{1}{8}\log_{15}(7)\color{black} - \color{blue} \left(\underbrace{\log_{15}(y^2)}_{\text{Same as }2\log_{15}(y)} + \log_{15}(x) \right) \color{black}= \tfrac{1}{8}\log_{15}(7) - 2\log_{15}(y) - \log_{15}(x)}$

Re: 8 problems that I can't understand after trying for days and looking examples onl

abender, I am just letting you know that I took the test and I nailed it! Never did I do a test better than this one. And you helped me a lot, those problems that you helped me solve and understand, were in the test and you are the reason that I knew how to do them and do them correctly. Thank you again! Happy holidays!