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

## 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
$\log_b x(\sqrt{x})$

## 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:

Mark44
Mentor

## 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?
$$log_b(x \sqrt{x})$$

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?

You're not solving for anything - you are evaluating this expression.

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

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

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

#### Attachments

• maths.jpg
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I assume you mean $log$$_{b}$$x$$\sqrt{x}$?

Try to get it so you have $klog$$_{b}$$x$, $k$ $constant$

Last edited:
eumyang
Homework Helper
Try to rewrite the expression
$\log_b(x \sqrt{x})$
as a single power, ie.
$\log_b(x^a)$
for some number a. You'll then need the power property of logarithms.

Try to rewrite the expression
$\log_b(x \sqrt{x})$
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
$\log_b x( \sqrt{x})$

that would be simple, but its
$\log_b x( \sqrt{x})$

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

And $x^{a}$$x^{b}$ = $x^{a+b}$

Mentallic
Homework Helper
that would be simple, but its
$\log_b x( \sqrt{x})$

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

$$(\log_bx)(\sqrt{x})$$

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 $\sqrt{x}$

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

$$(\log_bx)(\sqrt{x})$$

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 $\sqrt{x}$

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:
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 $log_{b}x$ = 0.8, so sub that in.

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
Homework Helper
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 $$log_b(x\sqrt{x})$$ as opposed to $$(log_bx)\cdot \sqrt{x}$$

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

Ok so it IS meant to be $$log_b(x\sqrt{x})$$ as opposed to $$(log_bx)\cdot \sqrt{x}$$

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
Mentor
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
Homework Helper
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!