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

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

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
Scoobiegal

Posts: 1
Joined: Fri Aug 07, 2009 1:43 am

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

As currently formatted, the equation is as follows:

. . . . .$10^x\, +\, \frac{10^{-x}}{10^x}\, -\, 10^{-x}\, =\, 5$

I have a feeling you meant something more like:

. . . . .$\frac{10^x\, +\, 10^{-x}}{10^x\, -\, 10^{-x}}\, =\, 5$

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:

. . . . .$10^x\, +\, \frac{1}{10^x}\, =\, 5\times 10^x\, -\, \frac{5}{10^x}$

Then multiply through by $10^x$, combine like terms, and rearrange to get:

. . . . .$\frac{3}{2}\, =\, 10^{2x}$

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

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!

stapel_eliz

Posts: 1804
Joined: Mon Dec 08, 2008 4:22 pm