# Thread: Calculus 1 - Problem

1. ## Calculus 1 - Problem

Hello, its my first question here. I need a help with a problem:
i have this equality 2logx = 4-log(x/10)

I did this:

2log10 x = 4 - log10 X - log10 10 (Log properties(log(X/Y) = logX - logY))
log10 x2 = 4 - log10 X - 1 (Log10 = 1)
log10 x2 = 3 - log10 X
log10 x2 + log10 X = 3
log10X3 = 3 (Log properties (log(XY) = logX + logY))
(log10X3)10 = 103 (Cancel the logarithm)
X3 = 103
X = 3√ 103

X = 10 (THIS IS MY FIRST SOLUTION)

I don't know if this is a right solution, but i found another solution later, doing again this:

2logx = 4-log(x/10)
log10X2 - log10 (X/10) = 4
(log10 (X3/10)) = 4
(log10(X3/10))10 = 104
X3/10 = 104 * 10
X3 = 3√ 105

X = 10
3√ 100 (SECOND SOLUTION)

What of the 2 solutions is correct? This is an exam, i don't know what to do :'(.
Thanks for all guys.

2. Originally Posted by LordDave
Hello, its my first question here. I need a help with a problem:
i have this equality 2logx = 4-log(x/10)

I did this:

2log10 x = 4 - log10 X - log10 10 (Log properties(log(X/Y) = logX - logY))
log10 x2 = 4 - log10 X - 1 (Log10 = 1)
log10 x2 = 3 - log10 X
log10 x2 + log10 X = 3
log10X3 = 3 (Log properties (log(XY) = logX + logY))
(log10X3)10 = 103 (Cancel the logarithm)
X3 = 103
X = 3√ 103

X = 10 (THIS IS MY FIRST SOLUTION)

I don't know if this is a right solution, but i found another solution later, doing again this:

2logx = 4-log(x/10)
log10X2 - log10 (X/10) = 4
(log10 (X3/10)) = 4
(log10(X3/10))10 = 104
X3/10 = 104 * 10
X3 = 3√ 105

X =
103√ 100 (SECOND SOLUTION)

What of the 2 solutions is correct? This is an exam, i don't know what to do
Did your instructor give you permission to seek outside help on this exam?

The expressions highlighted in red are not correct. In some cases, maybe you were thinking one thing, but typed something else? The other cases are wrong.

The expression in blue can be simplified.

You can test these candidates for x, by substituting them into the original equation. Evaluate, to see whether either of them yields a true statement.

Also, the notation log(x) means log base-10, so it's not necessary to type the base, in your expressions.

Also also, the symbols x and X are not interchangeable (they represent different things). Pick one symbol, and use it consistently throughout.

3. ## asd

Originally Posted by mmm4444bot
Did your instructor give you permission to seek outside help on this exam?

The expressions highlighted in red are not correct. In some cases, maybe you were thinking one thing, but typed something else? The other cases are wrong.

The expression in blue can be simplified.

You can test these candidates for x, by substituting them into the original equation. Evaluate, to see whether either of them yields a true statement.

Also, the notation log(x) means log base-10, so it's not necessary to type the base, in your expressions.

Also also, the symbols x and X are not interchangeable (they represent different things). Pick one symbol, and use it consistently throughout.

I did the exam with the first solution, i don't know if it's wrong, maybe yes.
I have only question, 2log10 x = 4 - logX - log10 , the correct solution here was 2logx = 4 - (logX - log10), right? Changing the - symbols.

Again thanks, and sorry for the different variables, i didn't change all before posting.

4. That's not a solution. And you still have x and X in the same equation. Part of success is being careful and consistent.

Can you solve this equation? 2Y = 4 - (Y - log10)

5. Originally Posted by tkhunny
That's not a solution. And you still have x and X in the same equation. Part of success is being careful and consistent.

Can you solve this equation? 2Y = 4 - (Y - log10)

The procedure i meant.

2log10 x = 4 - logx - log10 , the correct PROCEDURE here was 2logx = 4 - (logx - log10)
Is correct or no?

Can you solve this equation? 2Y = 4 - (Y - log10)
2Y = 4 - Y + log10
2Y = 4 - Y + 1
2Y = 5 - Y
2Y + Y = 5
3Y = 5
Y = 5/3, no?

6. Originally Posted by LordDave
I have only question, 2log10 x = 4 - logX - log10 … the correct solution here was 2logx = 4 - (logX - log10), right?
This is correct:

2 logx = 4 - (logx - log10)

This is not correct:

2 logx = 4 - logx - log10

7. I'm a little nervous about log(x) in this problem. In a calculus course, "log(x)" normally means Base e. Thus, it is difficult to pronounce "correct" without knowing the intent of the question. BTW, this is an Algebra Problem. Please get your algebra skills up to speed. Calculus will be very difficult without them.

8. Originally Posted by tkhunny
In a calculus course, "log(x)" normally means Base e.
Really? When did this happen?

The calculus courses I've seen (five colleges) used "ln" for Base-e, but it's been nearly five years since I've been personally involved.

I do know that the calculus and precalculus courses at my current employer use "ln" for Base-e.

9. Fair enough. Suffice it to say that there is some variation out there. Read your text carefully and be sure you know what it mean when it says "log(x)".