Replace 5^x with y and solve the equation for y. Then solve 5^x = y for the found value of y.

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- Jun 22nd 2011, 06:09 AM #1

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- Jun 22nd 2011, 06:17 AM #2

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- Jun 22nd 2011, 06:22 AM #3

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- Jun 22nd 2011, 06:23 AM #4

- Jun 22nd 2011, 06:31 AM #5

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## Re: Log question

fine....

I am sure it is 100% wrong, but since you insist.....

2(5^x+1) = 1 + 3/y

2(5^x+1) = y + 3/y

2y (5^x+1) = y+3

y( 5^x+1) = 3

5^2x+1 = 3 (I have a feeling I messed up bigtime, but I'll continue for your sake...)

(2x+1) log5 = log3

2x(log5) + log5 = log3

(stuck)

can you please help me now?

Thanks

- Jun 22nd 2011, 06:37 AM #6

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- Jun 22nd 2011, 06:38 AM #7

- Jun 22nd 2011, 06:48 AM #8

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## Re: Log question

ok so:

2(y+1) = 1 + 3/y

2y+2 = 1+ 3/y

2y+1 = 3/y

cross multiplying

2y^2+y=3

2y^2+y-3=0

aha! a quadratic equation!

solving it using -b+-sqrt(b^2-4ac)/2a gives either 1 or -1.5. keeping in mind that this is 5^x. x could be 0 or.... undefined...

unfortunately, this is not the answer (not even close for that matter) :-(

guys, please, would it kill you to answer the question and show your steps...?

Thanks

- Jun 22nd 2011, 07:00 AM #9

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- Jun 22nd 2011, 07:11 AM #10

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- Jun 22nd 2011, 07:18 AM #11

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- Jun 22nd 2011, 07:21 AM #12

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- Jun 22nd 2011, 07:36 AM #13

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- Jun 22nd 2011, 08:37 AM #14

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## Re: Log question

Last question....

so I know that x is -0.317394....... but, how did you know that this value+1 is equal to log5(3). It that a log rule? please tell me your method.

Thanks for your help :-)

- Jun 22nd 2011, 08:54 AM #15

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## Re: Log question

I did not claim above that , though this is true. If you don't have a calculator that computes logarithms to an arbitrary base, you can use the following equality.

(1)

In particular,

. (2)

The mnemonic for (1) is transitivity: (a,b) and (b,c) gives you (a,c). For (2), the argument of ln() in the nominator becomes the argument of log() and the argument from the denominator becomes the base, which is located in a subscript below the log argument, just like the denominator is below the nominator.

Also, see Wikipedia.

Starting from , you could also take ln() of both sides to get , from where and .

Edit: I am not sure I addressed the question you asked. If not, feel free to rephrase it.