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

  • Thread starter Thread starter srizen
  • Start date Start date
AI Thread Summary
The discussion revolves around solving the expression log_b(x√[3]{x}) given that log_b(x) = 0.8. Participants clarify that the problem involves evaluating the logarithm rather than solving for x directly. The correct approach involves using logarithmic properties to combine terms, leading to the realization that the answer simplifies to 1.07. Confusion arose from the unclear presentation of the problem, which required participants to deduce the correct interpretation. Ultimately, the issue stemmed from miscommunication rather than a lack of understanding of logarithmic concepts.
srizen
Messages
14
Reaction score
0

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 \log_b x =0.8

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

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:
Physics news on Phys.org
srizen said:

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?
log_b(x \sqrt[3]{x})

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

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?
 
Mark44 said:
You're not solving for anything - you are evaluating this expression.

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

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

hmm, let me scan it, i don't know how to do all these math symbols
 

Attachments

  • maths.jpg
    maths.jpg
    5.2 KB · Views: 569
I assume you mean log_{b}x\sqrt[3]{x}?

Try to get it so you have klog_{b}x, k constant
 
Last edited:
Try to rewrite the expression
\log_b(x \sqrt[3]{x})
as a single power, ie.
\log_b(x^a)
for some number a. You'll then need the power property of logarithms.
 
eumyang said:
Try to rewrite the expression
\log_b(x \sqrt[3]{x})
as a single power, ie.
\log_b(x^a)
for some number a. You'll then need the power property of logarithms.

that would be simple, but its
\log_b x( \sqrt[3]{x})
 
srizen said:
that would be simple, but its
\log_b x( \sqrt[3]{x})

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

And x^{a}x^{b} = x^{a+b}
 
srizen said:
that would be simple, but its
\log_b x( \sqrt[3]{x})

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

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

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

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

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

yea that's my question, how do you solve for x

when ur given 2 variables?

b is not given :S

heres what I am doing so far:b0.8 = x

Logb x + Logb 3√x

Logb x + 1/3Logb x

0.8 + 1/3Logb x ?
 
Last edited:
  • #10
srizen said:
yea that's my question, how do you solve for x

when ur given 2 variables?

b is not given :S

heres what I am 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 log_{b}x = 0.8, so sub that in.
 
  • #11
Bread18 said:
That's right. You know log_{b}x = 0.8, so sub that in.

the only problem is that, the answer should be 1.07 =/
 
  • #12
srizen said:
yea that's my question, how do you solve for x

when ur given 2 variables?

b is not given :S

heres what I am 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 log_b(x\sqrt[3]{x}) as opposed to (log_bx)\cdot \sqrt[3]{x}

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

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
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
srizen said:
ohh, ok i see my mistake now :S

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

Mark44 said:
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!
 
Back
Top