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"Im stuck on an equation. Anyone good at Logs? [Q] Solve: Log 2 X + Log 4 X = 2 I got to the next bit with the change of base rule: ( Log 10 ..."
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Old 07-05-2007, 15:46   #1 (permalink)
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Default Logarithm Question

Im stuck on an equation. Anyone good at Logs?

[Q] Solve: Log2X + Log4X = 2

I got to the next bit with the change of base rule:

(Log10X /
Log102) + (Log10X / Log104) = 2

Now Im stuck.



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Old 07-05-2007, 15:48   #2 (permalink)
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you can write the sum of two logs as the multiplication of the 2x and the 4x.

taking the antilog of both sides

(2x)*(4x) = antilog of 2
8x^2 = 100
x = 3.5355

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Old 07-05-2007, 15:56   #3 (permalink)
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Aha....so the change of base was going down the wrong way.

Cheers for that voodoo!

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Old 07-05-2007, 15:59   #4 (permalink)
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no worries!

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Old 09-05-2007, 23:49   #5 (permalink)
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Right, Voodoo or anyone else good at le mathmetique, I am in another Log predicament.

Question:

"Solve the simulatanoues equations leaving your answer as an exact fraction":

[E1] 8Y = 42X+3
[E2] Log2Y = log2X + 4

What voodoo stated before, taking the anitlogs: I dont know if this is the right direction to take but, I took the anti logs of [E2]. That left me with:

Y = X + 4

Then I subsituted this "X + 4" as "Y" into [E1] so I got:

8X+4 = 42X+3

Therefore this gives:

(84)(X4) = 42X+3


And Now Im Stuck. Help

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Old 09-05-2007, 23:52   #6 (permalink)
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Oh bluddy hell, the indices didnt come out right.!!

Leave it peeps.

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Old 09-05-2007, 23:53   #7 (permalink)
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from wat i am seeing cant u just take the X's on one side and solve for X?????
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Old 09-05-2007, 23:56   #8 (permalink)
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Originally Posted by AmericanBoi
from wat i am seeing cant u just take the X's on one side and solve for X?????
No, theres one problem:

In [E1] the "Y" is the power raised to 8.

and in [E2] "2X + 3" is the power raised to 4.

So it makes life difficult.

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Old 09-05-2007, 23:57   #9 (permalink)
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u can get an online calculator to do the powers of such equations if u dont have to show the work use that. but i am still confused as to how the original equation looks like can u put up a pic i can help u prolly.
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Old 10-05-2007, 00:14   #10 (permalink)
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Help me please.

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Old 10-05-2007, 00:15   #11 (permalink)
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The working under the [E1] and [E2] is mine. I dunno if that is right.

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Old 10-05-2007, 00:17   #12 (permalink)
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are u on msn?
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Old 10-05-2007, 00:18   #13 (permalink)
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LOL I'm keeping out of this one.

I'm glad I didn't take maths any further!

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Old 10-05-2007, 00:21   #14 (permalink)
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shit man i wasnt allowed log tables in ma exams i dont know how i solved log problems, but i was allowed calculators
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Old 10-05-2007, 00:24   #15 (permalink)
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We touched on this in college it was a nightmare to do!

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