solve for x algebraically: 10^x+10^-x/10^x-10^-x =5  TOPIC_SOLVED

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solve for x algebraically: 10^x+10^-x/10^x-10^-x =5

Postby Scoobiegal on Fri Aug 07, 2009 1:52 am

Can someone please show me how to work this problem? I am not really understanding it. The book gives the following answer =1/2 log 3/2

10^x+10^-x/10^x-10^-x =5
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Postby stapel_eliz on Fri Aug 07, 2009 12:42 pm

Scoobiegal wrote:10^x+10^-x/10^x-10^-x =5

As currently formatted, the equation is as follows:

. . . . .

Was this what you meant, or did you intend something else? :?:

I have a feeling you meant something more like:

. . . . .

If so, then a good first step in solving this exponential equation might be to multiply through to clear the (obvious) denominator, and then convert negative exponents to positive ones:

. . . . .

Then multiply through by , combine like terms, and rearrange to get:

. . . . .

Then take logs, etc, etc. But if you meant something else, then the above probably does not apply. :oops:

When you reply with correction or confirmation, please include a clear listing of what you have tried so far, so we can "see" where you are getting stuck. Thank you! :D
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