Quick Question.

[WTF?]

I have these two problems that i'm confused on how to do... :?

log85 - log17 + .5log25

and,
Solve for x: log x= 4 log2 +log3.

:?

 

[stapel]

1) What are the instructions?

2) It is already solved for x.

Edit: Well, it was solved for x, before you edited your post and changed the equation.

Eliz.

 

[WTF?]

The first one, it says to simplify. The second one, it says to solve for x. :?

 

[soroban]

Hello, WTF!

There must have been some instructions . . . . Simplify?

\(\displaystyle \log(85)\,-\,\log(17)\,+\,0.5\log(25)\)

We have: \(\displaystyle \;\log\left(\frac{85}{17}\right)\,+\,\log(25^{\frac{1}{2}})\;=\;\log(5)\,+\,\log(5)\;=\;2\cdot\log(5)\;=\;\log(5^2)\;=\;\log(25)\)

Solve for \(\displaystyle x:\;\; x\:=\:4\cdot\log(2)\,+\,\log(3)\)

It's already solved for \(\displaystyle x\) . . .

\(\displaystyle x\;=\;\log(2^4)\,+\,\log(3)\;=\;\log(16)\,+\,\log(3)\;=\;\log(16\cdot3)\;=\;\log(48)\)

 

[wjm11]

log x= 4 log2 +log3.

Soroban has already solved it for you, although I think he misread "log x" as being just x in the problem statement. Your final solution is

log x = log 48

or

x = 48

 

[WTF?]

ah, ok. Thanks.

 

[soroban]

Soroban has already solved it for you, although I think he misread "log x" as being just x in the problem statement.

Yes, I certainly did! . . . *blush*

 

[stapel]

Soroban...misread "log x" as being just x...

No. When Soroban replied, the equation read "x=". WTF? was editing the question while Soroban was typing up his reply, changing the equation to "log(x)".

This is a good example of why corrections are often more usefully made (or at least pointed out) in follow-up replies, rather than in "edits", which are often overlooked.

Eliz.

 

[WTF?]

Hello, I'm in need of another push. :!:

6^(n-5) = 400

Hm, don't know how to solve for n. I tried setting it up as n-5 log6 =log400 but no luck.

 

[stapel]

I'm in need of another push.

6^(n-5) = 400

I'm sorry, but I must be missing the connection. How did you get "6^n-5" from "log(x) = 4 log(2) + log(3)"?

Thank you.

Eliz.

 

[WTF?]

Hm, I'm confused now. If "n-5" is the power, shouldn't it go 'behind' the 6 given the correspondent property?

Can somebody please help me on how to set it up?

 

[soroban]

Hello, WTF?!

Hello, I'm in need of another push.
\(\displaystyle \;\;\;6^{n-5}\:=\:400\)

Hm, don't know how to solve for n.
I tried setting it up as: \(\displaystyle \,(n-5)\log6\:=\:\log400\) ... but no luck.
. . . Really? \(\displaystyle \;\)This should have worked!

You had: \(\displaystyle \,(n-5)\log(6)\:=\:\log(400)\)

\(\displaystyle \;\;\)Then: \(\displaystyle \,n\,-\,5\:=\;\frac{\log(400)}{\log(6)}\;\;\Rightarrow\;\;n\:=\:\frac{\log(400)}{\log(6)}\,+\,5\:\approx\:8.344\)

 

[WTF?]

Oh, thanks. I got it.

I have another one..

e^(x-1) = 25


I try to set it up as:

x-1 ln = ln25

Where to from there?

 

[WTF?]

also guys

If a graph has a y-intercept of 1 is it logarithmic.?

Or if it has no x-intercept is it exponential?

If it has a range of all real numbers it's exponential right?

 

[stapel]

I have another one...

Please post new questions as new threads, rather than as replies to old threads, where they tend to be overlooked.

Thank you for your consideration.

Eliz.

 

[WTF?]

Would this not be spam?

 

[stapel]

Would this not be spam?

Posting separate questions as separate posts is not "spam". Posting ads is spam.

Eliz.