thought i could do it... need sleep thats all

given that log r X = -5, log r Y = 2 and log r Z = 9, evaluate the expression

log r ( r^4 x-1 y^5 / 3[sqroot: x] )

Results 1 to 14 of 14

- Jun 3rd 2007, 08:57 PM #1

- Jun 3rd 2007, 09:07 PM #2

- Jun 3rd 2007, 09:19 PM #3
i think you meant cube root of x

$\displaystyle \log_r \left( \frac {r^4 x^{-1} y^5}{ \sqrt [3] {x}} \right)$

$\displaystyle = \log_r \left( \frac {r^4 y^5}{x \sqrt [3] {x}} \right)$ .......since $\displaystyle x^{-1} = \frac {1}{x}$

$\displaystyle = \log_r \left( r^4 y^5 \right) - \log_r \left( x^{ \frac {4}{3}} \right)$ ......since $\displaystyle \log_a \left( \frac {x}{y} \right) = \log_a x - \log_a y$

$\displaystyle = \log_r r^4 + \log_r y^5 - \log_r x^{ \frac {4}{3}}$ .....since $\displaystyle \log_a xy = \log_a x + \log_a y$

$\displaystyle = 4 \log_r r + 5 \log_r y - \frac {4}{3} \log_r x$ .......since $\displaystyle \log_a x^n = n \log_a x$

$\displaystyle = 4(1) + 5(2) - \frac {4}{3}(-5)$ ...........since $\displaystyle \log_a a = 1$

$\displaystyle = 4 + 10 + \frac {20}{3}$

I leave the last step to you

EDIT: If you have any questions, say so. I subtly applied some of the laws of exponents, make sure you don't miss them

- Jun 3rd 2007, 09:27 PM #4
Hello,

I assume that there is a typo and that you mean:

$\displaystyle \log_r \left(\frac{z^4 \cdot x^{-1} \cdot y^5}{3 \cdot \sqrt{x}} \right)$ = $\displaystyle 4 \cdot \log_r(z) - \log_r(x) + 5 \cdot \log_r(y) - \frac{3}{2} \cdot \log_r(x)$ Now plug in the values you know:

$\displaystyle = 4 \cdot 9 - (-5) + 5 \cdot 2 - \frac{3}{2} \cdot (-5) = 58.5$

- Jun 3rd 2007, 09:28 PM #5

- Jun 3rd 2007, 09:30 PM #6

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Hello, jessyc!

More questions . . .

Given that: .log r X = -5, log r Y = 2 and log r Z = 9, evaluate:

log r ( r^4 x^-1 y^5 / 3[sqroot: x] )

I assume we are given: .$\displaystyle \log_r(x) = -5,\;\log_r(Y) = 2,\;\log_r(Z) = 9$

. . and: .$\displaystyle \log_r\left(\frac{r^4\!\cdot\!X^{-1}\!\cdot\!Y^5}{3\sqrt{X}}\right) $

But I bet the denominator is: $\displaystyle \sqrt[3]{Z}$

- Jun 3rd 2007, 09:44 PM #7

- Jun 3rd 2007, 10:07 PM #8

- Jun 3rd 2007, 10:09 PM #9

- Jun 4th 2007, 03:47 AM #10

- Jun 4th 2007, 04:13 AM #11
Hello, jessyc,

I read your problem like this:

$\displaystyle \log_r \left(\frac{r^4 \cdot y^5}{x \cdot \sqrt[3]{z}} \right)= \frac{1}{x}$

I'm not quite sure what you want to do with this equation... So I'll try to simplify the LHS of it:

$\displaystyle 4 + 5\log_r(y)-\log_r(x) - \frac{1}{3} \cdot \log_r(z) = \frac{1}{x}$

Could it be that the RHS of this equation reads $\displaystyle \log_r \left(\frac{1}{x} \right) = -\log_r(x)$ ?

- Jun 4th 2007, 07:28 AM #12

- Jun 4th 2007, 07:59 AM #13
i think finding r exactly is pretty much impossible here, there are too many unknowns. as earboth suggested, they probably want us to prove that the left hand side is the same as the right hand side...or wait! maybe not! maybe we could find r....

pick up where earboth left off.

$\displaystyle 4 + 5\log_r(y)-\log_r(x) - \frac{1}{3} \cdot \log_r(z) = \frac{1}{x}$

$\displaystyle \Rightarrow 4 + 5 (2) - (-5) - \frac {1}{3} 9 = \frac {1}{x}$

$\displaystyle \Rightarrow 16 = \frac {1}{x}$

$\displaystyle \Rightarrow x = \frac {1}{16}$

Now we are told that $\displaystyle \log_r x = -5$

$\displaystyle \Rightarrow \log_r \left( \frac {1}{16} \right) = -5$

$\displaystyle \Rightarrow r^{-5} = \frac {1}{16}$ .......since, if $\displaystyle \log_a b = c \Rightarrow a^c = b$

$\displaystyle \Rightarrow r^5 = 16$

$\displaystyle \Rightarrow r = \sqrt [5] {16}$

and you can plug that into your calculator to see what it is

- Jun 4th 2007, 09:21 AM #14

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Hello, jessyc!

I assume I've incorporated the corrections . . .

Given: .$\displaystyle \log_r(X) = -5,\;\;\log_r(Y) = 2,\;\;\log_r(Z) = 9$

. . Evaluate: .$\displaystyle \log_r\!\left(\frac{r^4\!\cdot\!X^{-1}\!\cdot\!Y^5}{\sqrt[3]{Z}}\right) $

We have: .$\displaystyle \log_r\!\left(r^4\!\cdot\!X^{-1}\!\cdot\!Y^5\right) - \log_r\!\left(\sqrt[3]{Z}\right) $

. . $\displaystyle = \;\log_r\!\left(r^4\right) + \log_r\!\left(X^{-1}\right) + \log_r\!\left(Y^5\right) - \log_r\!\left(Z^{\frac{1}{3}}\right)$

. . $\displaystyle = \;4\!\cdot\!\underbrace{\log_r(r)}_{\downarrow} \,-\,1\!\cdot\!\underbrace{\log_r(X)}_{\downarrow} \,+\, 5\!\cdot\!\underbrace{\log_r(Y)}_{\downarrow} \,-\, \frac{1}{3}\!\cdot\!\underbrace{\log_r(Z)}_{\downa rrow}$

. . $\displaystyle = \;\;\;4\cdot(1) \;\;-\;\; 1\cdot(\text{-}5) \;\;\;+\;\;\; 5\cdot(2) \;\;\;\,-\,\;\;\; \frac{1}{3}\cdot(9)$

Answer: .$\displaystyle 4 + 5 + 10 - 3 \;=\;\boxed{16} $