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Old 08-03-08, 05:03 PM   #1
phantomcow2
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Logarithm



To solve for x I thought I would rewrite the equation in exponential form:

3^(2-x) = 3

Anything raised to the power of 1 is itself. So, x must equal 1. Because 3^(2-1) = 3

Why is this incorrect?
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Old 08-03-08, 05:07 PM   #2
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Originally Posted by phantomcow2 View Post


To solve for x I thought I would rewrite the equation in exponential form:

3^(2-x) = 3

Anything raised to the power of 1 is itself. So, x must equal 1. Because 3^(2-1) = 3

Why is this incorrect?

You're lucky so many people here like to show off their mad math skillz, otherwise you'd have spent a fortune on tutoring by now.
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Old 08-03-08, 05:15 PM   #3
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log_3(2 - x) = 3

3^log_3(2 - x) = 3^3

2 - x = 9

x = -7

Last edited by apricissimus; 08-03-08 at 05:19 PM.
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Old 08-03-08, 05:41 PM   #4
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I actually solved that a short bit ago. Thanks though.
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Old 08-03-08, 05:55 PM   #5
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Uh...3³ is not 9. It's 27.
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Old 08-03-08, 05:56 PM   #6
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Uh...3³ is not 9. It's 27.
I now realize that. THanks
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Old 08-03-08, 06:02 PM   #7
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Okay okay, now how about this:


The best I can do is rewrite it as .5xLog(.7) = 5log(1-x)
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Old 08-03-08, 06:22 PM   #8
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Is this your math hw?
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Old 08-03-08, 06:28 PM   #9
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Yes, and let me write my disclaimer now:

I seek guidance only, not the answer.
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Old 08-03-08, 06:31 PM   #10
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Just plug it into your calculator's numeric solver and be done with it.
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Old 08-03-08, 06:31 PM   #11
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That's not correct. You did the left side of the equation correctly, but you got things backwards on the right side.

Remember, log(x^y) = y* log(x)
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Old 08-03-08, 06:33 PM   #12
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Just plug it into your calculator's numeric solver and be done with it.
That's what made me mathematically handicapped in the first place!
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Old 08-03-08, 06:36 PM   #13
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Uh...3³ is not 9. It's 27.
*slaps palm on forehead*

I'm actually a math editor in real life. Minus a million points for me.

Last edited by apricissimus; 08-03-08 at 06:40 PM.
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Old 08-03-08, 06:39 PM   #14
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Oh, so I should have

.5xlog(.7) = 1-xlog(5)

then divide by log5 on both sides to begin isolation of the variable?
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Old 08-03-08, 06:45 PM   #15
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Oh, so I should have

.5xlog(.7) = 1-xlog(5)

then divide by log5 on both sides to begin isolation of the variable?
Remember that it's (1-x). Make sure you keep your parentheses arranged properly.
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Old 08-03-08, 06:56 PM   #16
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Yes, and let me write my disclaimer now:

I seek guidance only, not the answer.
I'm not faulting you for it. It's a shame more people aren't this proactive about their education.

Last edited by JaRow; 08-03-08 at 07:07 PM. Reason: bad spelling mistake
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Old 08-03-08, 07:04 PM   #17
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Okay okay, now how about this:


The best I can do is rewrite it as .5xLog(.7) = 5log(1-x)
By some algebraic manipulation, you can rewrite the equation as (35/2)^x = 5^2. (Try to do it yourself.)

From here take the log (to a certain base b) of both sides, and use the change of base formula, log_b(x) = log_a(x)/log_a(b) (or ln(x)/ln(b) if you like natural logarithms. And who doesn't?).
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Old 08-03-08, 07:17 PM   #18
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By some algebraic manipulation, you can rewrite the equation as (35/2)^x = 5^2. (Try to do it yourself.)

From here take the log (to a certain base b) of both sides, and use the change of base formula, log_b(x) = log_a(x)/log_a(b) (or ln(x)/ln(b) if you like natural logarithms. And who doesn't?).
And as usual, it's the algebraic manipulation which sinks me.
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Old 08-03-08, 07:23 PM   #19
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And as usual, it's the algebraic manipulation which sinks me.
Here's the first couple of steps:

Rewrite as

(0.7^x)^1/2 = 5*(5^(-x))

Then square both sides...

0.7^x = (5^2)*(5^(-2x))

etc.
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