If Logb(x) = 0.8, solve for Logb(x)(x^1/3)

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  • #1
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Homework Statement



i got a question on a quiz, but i really did not know how to solve it. the question that was stated was:
if [itex]\log_b x =0.8[/itex]

Solve for
[itex]\log_b x(\sqrt[3]{x})[/itex]


The Attempt at a Solution



basically what i attempted to do was
b0.8 = x

Logb x + Logb 3√x

Logb x + 1/3Logb x

0.8 + 1/3Logb x ?


EDIT:
picture attachment
formatting
 
Last edited:

Answers and Replies

  • #2
35,407
7,278

Homework Statement



i got a question on a quiz, but i really did not know how to solve it. the question that was stated was:
if Logbx = 0.8

Solve for
Logbx(3√x)
You're not solving for anything - you are evaluating this expression.

In any case, is that what you're working with?
[tex]log_b(x \sqrt[3]{x})[/tex]

If so, that's different from what you wrote.

The Attempt at a Solution



basically what i attempted to do was

Make Logbx = 0.8
into b0.8 = x
then i didn't know what to do ;S.

what steps do i need to do?
how do i solve for b/x?
 
  • #3
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You're not solving for anything - you are evaluating this expression.

In any case, is that what you're working with?
[tex]log_b(x \sqrt[3]{x})[/tex]

If so, that's different from what you wrote.

hmm, let me scan it, i dont know how to do all these math symbols
 

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  • #4
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I assume you mean [itex]log[/itex][itex]_{b}[/itex][itex]x[/itex][itex]\sqrt[3]{x}[/itex]?

Try to get it so you have [itex]klog[/itex][itex]_{b}[/itex][itex]x[/itex], [itex]k[/itex] [itex]constant[/itex]
 
Last edited:
  • #5
eumyang
Homework Helper
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Try to rewrite the expression
[itex]\log_b(x \sqrt[3]{x})[/itex]
as a single power, ie.
[itex]\log_b(x^a)[/itex]
for some number a. You'll then need the power property of logarithms.
 
  • #6
14
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Try to rewrite the expression
[itex]\log_b(x \sqrt[3]{x})[/itex]
as a single power, ie.
[itex]\log_b(x^a)[/itex]
for some number a. You'll then need the power property of logarithms.

that would be simple, but its
[itex]\log_b x( \sqrt[3]{x})[/itex]
 
  • #7
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that would be simple, but its
[itex]\log_b x( \sqrt[3]{x})[/itex]

Use the relationship
[itex]x[/itex][itex]^{\frac{1}{a}}[/itex] = [itex]\sqrt[a]{x}[/itex]

And [itex]x^{a}[/itex][itex]x^{b}[/itex] = [itex]x^{a+b}[/itex]
 
  • #8
Mentallic
Homework Helper
3,798
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that would be simple, but its
[itex]\log_b x( \sqrt[3]{x})[/itex]

If it's supposed to be in that form as the question asks, which I suppose should mean

[tex](\log_bx)(\sqrt[3]{x})[/tex]

Then simply plug in 0.8 into the first bracket since you're already given that, and solve for x in [itex]\log_bx=0.8[/itex] to find [itex]\sqrt[3]{x}[/itex]
 
  • #9
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If it's supposed to be in that form as the question asks, which I suppose should mean

[tex](\log_bx)(\sqrt[3]{x})[/tex]

Then simply plug in 0.8 into the first bracket since you're already given that, and solve for x in [itex]\log_bx=0.8[/itex] to find [itex]\sqrt[3]{x}[/itex]

yea thats my question, how do you solve for x

when ur given 2 variables?

b is not given :S

heres what im doing so far:


b0.8 = x

Logb x + Logb 3√x

Logb x + 1/3Logb x

0.8 + 1/3Logb x ?
 
Last edited:
  • #10
100
0
yea thats my question, how do you solve for x

when ur given 2 variables?

b is not given :S

heres what im doing so far:


b0.8 = x

Logb x + Logb 3√x

Logb x + 1/3Logb x

0.8 + 1/3Logb x ?

That's right. You know [itex]log_{b}x[/itex] = 0.8, so sub that in.
 
  • #11
14
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That's right. You know [itex]log_{b}x[/itex] = 0.8, so sub that in.

the only problem is that, the answer should be 1.07 =/
 
  • #12
Mentallic
Homework Helper
3,798
94
yea thats my question, how do you solve for x

when ur given 2 variables?

b is not given :S

heres what im doing so far:


b0.8 = x

Logb x + Logb 3√x

Logb x + 1/3Logb x

0.8 + 1/3Logb x ?

Ok so it IS meant to be [tex]log_b(x\sqrt[3]{x})[/tex] as opposed to [tex](log_bx)\cdot \sqrt[3]{x}[/tex]

Yes, the answer is 1.07, what's the problem? 0.8 + 1/3*0.8 = ?
 
  • #13
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Ok so it IS meant to be [tex]log_b(x\sqrt[3]{x})[/tex] as opposed to [tex](log_bx)\cdot \sqrt[3]{x}[/tex]

Yes, the answer is 1.07, what's the problem? 0.8 + 1/3*0.8 = ?

ohh, ok i see my mistake now :S

this is such an easy problem, i was jus over thinking >_<
 
  • #14
35,407
7,278
Whoever wrote this problem did not communicate the problem in a clear way. The only way we could understand what was written was to reverse-engineer the solution. The problem should have been written so that it was clear what the argument of the log function was.
 
  • #15
Mentallic
Homework Helper
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ohh, ok i see my mistake now :S

this is such an easy problem, i was jus over thinking >_<
Yep :wink:

Whoever wrote this problem did not communicate the problem in a clear way. The only way we could understand what was written was to reverse-engineer the solution. The problem should have been written so that it was clear what the argument of the log function was.
I was about to ask for the solution if the OP didn't already provide it. This confusion was unnecessarily getting out of hand...

edit: the answer was written in the attachment. To be honest, it looked like effort to try and understand what was happening on that page so I gave up!
 

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