Exponential equation  TOPIC_SOLVED

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Exponential equation

Postby winkylocc on Thu Mar 31, 2011 3:39 am

I have a problem I'm posting here. I already have the answer that Im posting as well along with the steps that the instructor wrote next to my incorrect answer. I need help with one of the steps that completely lost me.
Here is the problem:

5 2x+1 = 7 3-x

The steps follow:

2x+1 = (3-x) log 5 7
What he does next with the (3-x) is what really throws me:
2x+1 = 3 log5 7 - x log 5 7

2x + x log 5 7 = 3 log 5 7 - 1

x(2 + log 5 7) = 3 log 5 7 - 1

The answer is:

x = 3 log 5 7 -1
    2 + log 5 7

Again, I was with him until the right side of the second step where things started happening that I cant figure out. Thanks for the help.
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Postby stapel_eliz on Thu Mar 31, 2011 11:08 am

winkylocc wrote:5 2x+1 = 7 3-x

2x+1 = (3-x) log 5 7

What he does next with the (3-x) is what really throws me:

2x+1 = 3 log5 7 - x log 5 7

He just distributed, like this:

. . . . .(3 - x)y = 3(y) - x(y) = 3y - xy

:wink:
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Re: Exponential equation

Postby winkylocc on Thu Mar 31, 2011 4:03 pm

How will I know to distribute in a problem like that.
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  TOPIC_SOLVED

Postby stapel_eliz on Thu Mar 31, 2011 10:34 pm

winkylocc wrote:How will I know to distribute in a problem like that.

Sometimes distributing is helpful; sometimes factoring is helpful. In general, use the step(s) you need for isolating the variable. :wink:
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