You have like terms, so
becomes
add 8
divide by 35
natural logs, etc...
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Now, I suspect that there is a typo in your original problem. If you confirm this, then I will work out the quadratic.
HEY GUYS,
I'm working my way through an online course and it is impossible getting a phone call back from my teachers. This equation has been bothering me for days now and its holding me back.
4(2^x)+31(2^x)-8=0
The lesson i'm on is exponential and logarithmic equations, and i understand i'm supposed to find common bases' and apply logarithms, but for some reason this question is beyond me. Any help in the right direction would be much appreciated. THANKS
When you start with three terms, all on the same side of the equation.
When the middle coefficient (31) is the sum of two factors of the product of the other coefficients ("ac" = 4*-8 = -32 = 32 * -1; 32 + -1 = 31 = "b").
When your life is no longer your own. Oh wait, that's something else.