2^(x^2 - 35)= 4^x
How do I solve this? The x^2 is messing me up. So far I have:
ln(2^(x^2-35))=ln(4^x)
(x^2-35)ln 2= xln4
x^2ln 2 - 35ln2= xln4
x^2ln 2- xln4= 35ln2
x(xln2 - ln4)= 35ln2
I would have taken the log base 2 of each side, but ok, with what you have, ln4 is 2ln2. Then divide through by ln2.
here is an easy way to do that.
Notice the bases of the powers are 2 and 4, so let's change 4 to base 2
2^(x^2 - 35)= 4^x
2^(x^2 - 35)= 2^2x
then x^2 - 35 = 2x
x^2 - 2x - 35 = 0
(x-7)(x+5)=0
x=7 or x=-5
To solve the equation 2^(x^2 - 35) = 4^x, you can convert the right side of the equation to have the same base as the left side. Since both 2 and 4 are powers of 2, we can rewrite 4 as (2^2):
2^(x^2 - 35) = (2^2)^x
Using the properties of exponents, we can simplify the equation further:
2^(x^2 - 35) = 2^(2x)
Now that the bases are the same, we can equate the exponents:
x^2 - 35 = 2x
Next, rearrange the equation to have all the terms on one side:
x^2 - 2x - 35 = 0
Now we have a quadratic equation in terms of x. To solve this, we can factorize or use the quadratic formula. Let's factorize it:
(x-7)(x+5) = 0
By applying the zero product property, we can set each factor equal to zero:
x - 7 = 0 --> x = 7
or
x + 5 = 0 --> x = -5
So the potential solutions for x are x = 7 and x = -5.