How to solve ln(2^x+2)=2xln(2) ?
First attempt:
ln(2^x+2)=2xln(2)
ln(2^x+2)=ln(2^2x)
2^x+2=2^2x
2^x-2^2x+2=0
+33616 Half right!
If y = 2 then 2^x = 2 so x = 1 ok
But if y = -1 then you need 2^x = -1 and this is not possible in the real number domain. When x is -1, 2^-1 = 1/2 not -1.
2^x is always a positive number for x positive or negative.
So the only valid solution here is x = 1
Put each value of x back into the original equation to check.
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+33616 You are doing ok so far. Next step is to write 2^2x as (2^x)^2 then let y = 2^x so you have
y^2 - y - 2 = 0 (Taking each term in your last line to the other side first, so the y^2 term is positive)
This factors as
(y + 1)(y - 2) = 0
Can you take it from here?
.
+33616 Half right!
If y = 2 then 2^x = 2 so x = 1 ok
But if y = -1 then you need 2^x = -1 and this is not possible in the real number domain. When x is -1, 2^-1 = 1/2 not -1.
2^x is always a positive number for x positive or negative.
So the only valid solution here is x = 1
Put each value of x back into the original equation to check.
.
