1. lns + ln(x+2) =4
i get this down to x^2 +2x -e^4= zero. But I can't figure it out from there. Do I just use quadratic formula?
2. 3^(2x) + 3^(x+1) - 0
i get this down to x=((4-ln3)/(ln3)) * 1/3
but I can't figure out how to get the solution.
Thanks for any help in advance.
I'm assuming that you mean this:
Now, is the same as , and is the same as :
Now, let's substitute 3^x = t as javax suggested:
Hey, that looks like a quadratic equation! Solve for t:
or
But we're not done! We need to solve for x. So substitute 3^x back in for t:
or
Hmm, there's no x that would satisfy either equation. The base is positive and greater than 1, so the range of the function y = 3^x would be [0, ∞). So there is no solution, unless in the original equation the RHS equals some other number instead of 0.
01
I think I figured out those 2. Thanks guys.
Here's another confusing exponential. It involves multiplication by constants.
2*49^x +11*7^x +5= 0
I'm guessing I should start by making 49 a 7^2 and moving 5 over.
2*(7^2)^x +11*7^x = -5
Do I just divide the 2 and 11 out? Or something else? Thanks again.