Logarithm help

[brandonbot]

Hi i just have this equation i'm suppose to solve using logarithms however i'm a little stuck cause the one is raised to the power of x while the other is raised to the power of -x not sure how to proceed. Any help would be appreciated thanks

10^x-10^-x/2=8

 

[Deleted member 4993]

Hi i just have this equation i'm suppose to solve using logarithms however i'm a little stuck cause the one is raised to the power of x while the other is raised to the power of -x not sure how to proceed. Any help would be appreciated thanks

10^x-10^-x/2=8

Problem as posted is ambiguous.

Use some parentheses to indicate order of operation.

 

[brandonbot]

Sorry its 10 to the power of x -10 to the power of negative x all divided by 2 = 8 i have to use logarithms to solve for x sorry just not sure how to type it out using numbers

 

[srmichael]

Sorry its 10 to the power of x -10 to the power of negative x all divided by 2 = 8 i have to use logarithms to solve for x sorry just not sure how to type it out using numbers

You mean this?: \(\displaystyle \displaystyle \frac{10^{(x-10)^{-x}}}{2}=8\)

 

[brandonbot]

Sorry close but the terms on top are seperate its 10 to the power of x subtract 10 to the power of negative x

 

[pka]

Sorry close but the terms on top are seperate its 10 to the power of x subtract 10 to the power of negative x

So it is \(\displaystyle 10^x-10^{-x}=16\) [multiple by 2]
which is the same as \(\displaystyle 10^{2x}-16\cdot 10^{x}-1=0\). [multiply by \(\displaystyle 10^x\).

If you can solve \(\displaystyle y^2 - 16y - 1 = 0\) you can let \(\displaystyle y = 10^x\).