I don't know if this will show up properly, but do you mean
52x - 45x = 12 ?
How did you get log(5^x) = log(-4) ??
Could someone help me with this question?
5^(2x) - 4^(5^x) = 12
I got up to log (5^x)= log(-4)
but it isnt correct since it can't be negative.
5 answers
Yes
to get that i did:
(5^x)^2 - 4(5^x) =12
5^x(1-4) = 12
-3(5^x)= 12
log(5^x) = log(-4)
If you could show me the correct way to do it, that would be great! thanks!
to get that i did:
(5^x)^2 - 4(5^x) =12
5^x(1-4) = 12
-3(5^x)= 12
log(5^x) = log(-4)
If you could show me the correct way to do it, that would be great! thanks!
ah, but now you changed the question.
In the original you had 4 raised to the (5 raised to the x) in the middle term
Now you have 4(5^x)
This last version makes it easy, the first version would be a nightmare.
If it is
(5^x)^2 - 4(5^x) =12
then let's let y = 5^x
our equation now becomes
y^2 - 4y = 12 which is a nice quadratic
y^2 - 4y - 12 = 0
(y-6)(y+2) = 0
so y = 6 or y = -2
then 5^x = 6
log (5^x) = log6
xlog5 = log6
x = log6/log5 = 1.11328
or 5^x = -2
xlog5 = log(-2) which is undefined, thus no solution for that part
so x = appr. 1.113
In the original you had 4 raised to the (5 raised to the x) in the middle term
Now you have 4(5^x)
This last version makes it easy, the first version would be a nightmare.
If it is
(5^x)^2 - 4(5^x) =12
then let's let y = 5^x
our equation now becomes
y^2 - 4y = 12 which is a nice quadratic
y^2 - 4y - 12 = 0
(y-6)(y+2) = 0
so y = 6 or y = -2
then 5^x = 6
log (5^x) = log6
xlog5 = log6
x = log6/log5 = 1.11328
or 5^x = -2
xlog5 = log(-2) which is undefined, thus no solution for that part
so x = appr. 1.113
The original question was what i needed to solve, but with the way i changed it, is it correct?
My answer, if substituted into the second version of the equation, satisfies it.