Question 233589: I am really struggling to find the answer to this logarithmic equation that we have been set:
6*(3^x)+4*(3^-x)-14=0
Any help would be appreciated,
Many thanks, Trevor
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 6*(3^x)+4*(3^-x)-14=0
3^(-x) is the same as 1/3^x
your equation becomes:
6*(3^x) + 4*(1/3^x) - 14 = 0
multiply both sides of this equation by 3^x to get:
6 * (3^x) * (3^x) + 4 - 14 * (3^x) = 0
this becomes:
6 * (3^x)^2 - 14 * (3^x) + 4 = 0
Let y = 3^x and your equation becomes:
6*y^2 - 14*y + 4 = 0
divide both sides of this equation by 2 to get:
3*y^2 - 7*y + 2 = 0
You can factor this equation to get:
(3y-1) * (y-2) = 0
This makes:
3y-1 = 0 or y-2 = 0
3y-1 = 0 solves for y to get y = 1/3
y-2 = 0 solves for y to get y = 2
your answers so far are:
y = 1/3 or y = 2
since y = 3^x, this means that:
3^x = 1/3 or 3^x = 2
this is where the logarithms come in. up to this point it was straight algebra.
work with 3^x = 1/3 first.
take log of both sides to get:
log(3^x) = log(1/3)
this becomes:
x*log(3) = log(1/3)
divide both sides by log(3) to get:
x = log(1/3)/log(3)
this becomes:
x = -1
work with 3^x = 2 next.
take log of both sides to get:
log(3^x) = log(2)
this becomes:
x*log(3) = log(2)
divide both sides by log(3) to get:
x = log(2)/log(3)
this becomes:
x = .630929754
your answer are:
x = -1
or
x = .630929754
you need to confirm these answers are good by plugging into the original equation.
original equation is:
6*(3^x)+4*(3^-x)-14=0
if x = -1, this becomes:
6*(3^(-1))+4*(3^-(-1))-14=0
this becomes:
6/3 + 4*3 - 14 = 0 which becomes 2 + 12 - 14 = 0 which becomes 0 = 0 which is true so x = -1 is good.
if x = .630929754, this becomes:
6*(3^(.630929754))+4*(3^-(.630929754))-14=0
this becomes:
6*2 + 4*(1/2) - 14 = 0 which becomes 12 + 2 - 14 = 0 which becomes 0 = 0 which is true so x = .630929754 is good.
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