exponents they make me cry
okay so guys, i've stumbled upon some difficulty with the exponential functions and i'm in some need of help
2^(x^2 + 2x) = 2^(x + 6)
1/√2 = (1/2) ^ (8 / x)
would be much appreciated if the steps were shown and no, this is not homework.
November 12, 2012
12 Comments • Newest first
Why is everyone picking base 2 for the second part?
take the base as 1/√2, then you have (1/√2)^1 = ((1/√2)^2)^(8/x); 1 = 16/x.
You're quite lucky the first question has given you in the same base of 2.
In short, since the base are the same, you solve the exponent just like simpler algebra:
x^2+2x=x+6.
x^2-x-6=0
(x+3)(x-2)=0 x= -3 or 2
2.
When solving exponential functions equations, use logarithms or set the same base.
1/√2 = (1/2) ^ (8 / x)---> 2^(-1/2) = 2^(-8/x)
Now solve x in the exponents:
-1/2=-8/x
x=16
Oh god my next math class is going to suck terribly. I have only a faint recollection of this stuff.
@Fiercerain: why are you using logs in this?
2^something = 2^something else
x^2+2x=x+6
x^2+x-6 = 0
(x+3)(x-2) = 0.
x = -3 or 2.
(2)^-1/2= (2)^-8/x
-1/2=-8/x
-x=-16
x=16, yah!
[quote=Fiercerain]@Creashun: You might have to do change of log if you can't assign a base other than 10. To do this you can use this:
Log(value) / Log(base)
so if I wanted to do log(base of 3) (81) = 4 you would plug in:
Log(81) / Log(3) and you should get 4[/quote]
oh, many thank yous <:
@Creashun: You might have to do change of log if you can't assign a base other than 10. To do this you can use this:
Log(value) / Log(base)
so if I wanted to do log(base of 3) (81) = 4 you would plug in:
Log(81) / Log(3) and you should get 4
[quote=Fiercerain]for the first one, set both sides to log base of 2, and move everything to one side and factor.
For the second one, set both sides to log base of 1/2 and find out what x equals ;o
to make sure you understand the concept I'm suggesting, plug in log(base of 2)(2^3) and you should get 3 as an answer.
Let me know if you'd like me to sketch out the steps, but this pretty much sums up what you have to do.[/quote]
oh wait nevermind
this is the easiest part in the curriculum
o-o i came in thinking oh snap exponents be easy i can do this took one look at the question and cried
for the first one, set both sides to log base of 2, and move everything to one side and factor.
For the second one, set both sides to log base of 1/2 and find out what x equals ;o
to make sure you understand the concept I'm suggesting, plug in log(base of 2)(2^3) and you should get 3 as an answer.
Let me know if you'd like me to sketch out the steps, but this pretty much sums up what you have to do.
For the first one:
x^2+2x = x+6
Use the quadratic formula or factorization to solve for "x". Plug in the "x" value into the original equation. Voila!