Question 614596: Find the solution of the exponential equation..
3^2x-1=5
7^x/2= 5^1-x
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! first equation, as far as i can tell, is:
3^(2x-1) = 5
you really need parentheses to remove any doubt as to which part of the expression is in the exponent and which part isn't.
based on what i think it is, the solution would be:
take the log of both sides of the equation to get:
log(3^(2x-1)) = log(5)
this becomes:
(2x-1) * log(3) = log(5)
divide both sides of the equation by log(3) to get:
2x-1 = log(5)/log(3)
add 1 to both sides of the equation and then divide both sides of the equation by 2 to get:
x = (log(5)/log(3)+1)/2
this results in:
x = 1.232487
plug that value into the original equation to confirm the value is good.
3^(2x-1) = 5 becomes:
3^(2*1.232487 - 1) = 5 which becomes:
5 = 5, confirming the value for x is good.
answer is that x = 1.232487
logarithm properties that you are using are:
log(a^x) = x*log(a)
that's what allowed me to make log(3^(2x-1)) equivalent to (2x-1)*log(3).
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second equation, as far as i can tell, is:
7^(x/2) = 5^(1-x)
take the log of both sides of this equation to get:
log(7^(x/2)) = log(5^(1-x))
This becomes:
(x/2)*log(7) = (1-x)*log(5)
divide both sides of this equation by (1-x) and divide both sides of this equation by log(7) to get:
(x/2) / (1-x) = log(5) / log(7)
multiply both sides of this equation by 2*(1-x) to get:
x = 2*(1-x)*(log(5)/log(7)
simplify this equation by performing indicated operations to get:
x = 2*log(5)/log(7) - 2*x*log(5)/log(7)
add 2*x*log(5)/log(7) to both sides of this equation to get:
x + 2*x*log(5)/log(7) = 2*log(5)/log(7)
factor out the x on the left side of the equation to get:
x * (1 + 2*log(5)/log(7)) = 2*log(5)/log(7)
divide both sides of this equation by (1+2*log(5)/log(7)) to get:
x = (2*log(5)/log(7)) / (1 + 2*log(5)/log(7))
solve for x using your calculator to get:
x = .623235085
plug this value of x into your original equation to see if that equation holds true.
your original equation is:
7^(x/2) = 5^(1-x)
replace x with .623235085 to get:
7^(.623235085/2) = 5^(1-.623235085)
this results in:
7^(.311617542) = 5^(.376764915) which results in:
1.833780602 = 1.833780602
Since the results match, the value of x is good.
logarithm properties that you are using are:
log(a^x) = x*log(a)
that's what allowed me to say that:
log(7^(x/2)) = log(5^(1-x)) is equivalent to (x/2)*log(7) = (1-x)*log(5)
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