# Thread: log question (last one)

1. ## log question (last one)

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] )

2. Originally Posted by Jhevon
i assume you mean log to the base r when you write log r

what is that x - 1? should it be x^-1? please use parenthesis to make your question clear

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

edit, base of r , you where correct

3. Originally Posted by jessyc
given that log r X = -5, log r Y = 2 and log r Z = 9, evaluate the expression

logr ( {r^4} {x^-1} {y^5} / 3[sqroot: x] )
i think you meant cube root of x

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

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

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

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

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

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

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

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

4. Originally Posted by jessyc
logr ( {r^4} {x^-1} {y^5} / 3[sqroot: x] )

edit, base of r , you where correct
Hello,

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

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

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

5. Originally Posted by earboth
Hello,

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

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

$= 4 \cdot 9 - (-5) + 5 \cdot 2 - \frac{3}{2} \cdot (-5) = 58.5$
yeah, i was wondering about that r being there and no z

i don't think it would be $3 \cdot \sqrt {x}$ though, since you would end up with $log_r 3 + \frac {1}{2} \log_r x \neq \frac {3}{2} \log_r x$

6. 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: . $\log_r(x) = -5,\;\log_r(Y) = 2,\;\log_r(Z) = 9$

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

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

7. this is my work so far i'm rewriting the questions so they look better
this is proof of my work...

the pages are double sided... anyways thanks

edit: oh and its 2:45 AM got school at 9 have to get up at 8 love life (thanks for making it easyer guys )

8. Originally Posted by Soroban
Hello, jessyc!

More questions . . .

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

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

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

how would i do it?
...

9. Originally Posted by jessyc
how would i do it?
...
exactly the way you saw us do the last one. except at the second to last line, you would plug in the value for Z instead. you should try it and tell us your answer

10. it was late last night....

logr ({r^4}{y^5}/x{3rd root of z}) = 1/x

logr^4{y^5} = x * {3rd root of z}

logr^4 = x * {3rd root of z} / y^5

r^4 = 0.3250
(then just rais to the 4th power.)

this answer doesn't look right to me...

11. Originally Posted by jessyc
it was late last night....

logr ({r^4}{y^5}/x{3rd root of z}) = 1/x

logr^4{y^5} = x * {3rd root of z}

logr^4 = x * {3rd root of z} / y^5

r^4 = 0.3250
(then just rais to the 4th power.)

this answer doesn't look right to me...
Hello, jessyc,

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

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

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

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

12. sorry x,y,z have values,
im not sure what i was suppose to do with it either.
log r X = -5, log r Y = 2 and log r Z = 9,
are the values

i was assuming i was trying to find r...

13. Originally Posted by jessyc
sorry x,y,z have values,
im not sure what i was suppose to do with it either.
log r X = -5, log r Y = 2 and log r Z = 9,
are the values

i was assuming i was trying to find r...
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.

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

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

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

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

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

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

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

$\Rightarrow r^5 = 16$

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

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

14. Hello, jessyc!

I assume I've incorporated the corrections . . .

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

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

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

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

. . $= \;4\!\cdot\!\underbrace{\log_r(r)}_{\downarrow} \,-\,1\!\cdot\!\underbrace{\log_r(X)}_{\downarrow} \,+\, 5\!\cdot\!\underbrace{\log_r(Y)}_{\downarrow} \,-\, \frac{1}{3}\!\cdot\!\underbrace{\log_r(Z)}_{\downa rrow}$
. . $= \;\;\;4\cdot(1) \;\;-\;\; 1\cdot(\text{-}5) \;\;\;+\;\;\; 5\cdot(2) \;\;\;\,-\,\;\;\; \frac{1}{3}\cdot(9)$

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