# 8 problems that I can't understand after trying for days and looking examples online

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• December 17th 2012, 08:45 PM
Heisenberg
Re: 8 problems that I can't understand after trying for days and looking examples onl
Quote:

Originally Posted by abender
$f(x)=(x-2)^3$

$y=(x-2)^3$

Switching $x$ and $y$:
$x = (y-2)^3$

Now, to solve for $y$, raise both sides to the $\tfrac{1}{3}$ power:
$x^{\frac{1}{3}} = ((y-2)^3)^{\frac{1}{3}}$

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

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 $y^{2}x$?

Yes.
• December 17th 2012, 09:14 PM
abender
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) $\log_b(xy) = \log_b(x) + \log_b(y)$
2) $\log_b(\tfrac{x}{y}) = \log_b(x) - \log_b(y)$
3) $\log_b(x^n) = n\log_b(x)$

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

$\small{\log_{15}\frac{7^{\frac{1}{8}}}{y^2x} = \color{red}\log_{15}(7^{\frac{1}{8}}) \color{black} - \color{blue}\log_{15}(y^2x)\color{black} = \color{red}\frac{1}{8}\log_{15}(7)\color{black} - \color{blue} \left(\underbrace{\log_{15}(y^2)}_{\text{Same as }2\log_{15}(y)} + \log_{15}(x) \right) \color{black}= \tfrac{1}{8}\log_{15}(7) - 2\log_{15}(y) - \log_{15}(x)}$
• December 18th 2012, 10:55 AM
Heisenberg
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!
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