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

[srizen]

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

 

[Mark44]

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.

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?

 

[srizen]

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

 

[Bread18]

I assume you mean log_{b}x\sqrt[3]{x}?

Try to get it so you have klog_{b}x, k constant

 

[eumyang]

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.

 

[srizen]

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})

 

[Bread18]

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}

 

[Mentallic]

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}

 

[srizen]

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:b^0.8 = x

Log_b x + Log_b ³√x

Log_b x + 1/3Log_b x

0.8 + 1/3Log_b x ?

 

[Bread18]

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:


b^0.8 = x

Log_b x + Log_b ³√x

Log_b x + 1/3Log_b x

0.8 + 1/3Log_b x ?

That's right. You know log_{b}x = 0.8, so sub that in.

 

[srizen]

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 =/

 

[Mentallic]

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:b^0.8 = x

Log_b x + Log_b ³√x

Log_b x + 1/3Log_b x

0.8 + 1/3Log_b 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 = ?

 

[srizen]

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 >_<

 

[Mark44]

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.

 

[Mentallic]

ohh, ok i see my mistake now :S

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

Yep

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!