Solving for x, right? Gimme a sec.
For the first one I get x=log(2+-root3) - the log in base 10 of 2 plus or minus the square root of 3.
First off, take the 3 on the left hand side (LHS) inside the brackets and expand the square on the RHS to get:
3*10^(2x) + 6*10^x = 4*10^(2x) + 2*10^x + 1
then tidy that up and move everything to the RHs to get a quadratic equation in 10^x:
10^(2x) - 4*10^x + 1 =0
Then that has real solutions where 10^x = (4+-root12)/2 = 2+-root3
so x=log(2+-root3)
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