Question 843561: Solve each equation. Show all your work. Round your answers to four decimal places.
a. 63x = 104
b. log3(x + 2) + log3(4x – 1) = 5
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 63x = 104
divide both sides of this equation by 63 to get:
x = 104/63
log3(x+2) + log3(4x-1) = 5
use the log property of log(a) + log(b) = log(a*b) to get:
log3((x+2)*(4x-1)) = 5
simplify the expression within the log to get:
log3(4x^2 + 7x - 2) = 5
the basic property of logs states that logb(x) = y if and only if b^y = x
apply this property to the equation to get:
3^5 = 4x^2 + 7x - 2
simplify to get:
243 = 4x^2 + 7x - 2
subtract 243 from both sides of this equation to get:
0 = 4x^2 + 7x - 245
commute this equation to get:
4x^2 + 7x - 245 = 0
factor this equation by using the quadratic formula to get:
x = (-7 +/- sqrt(3969)) / 8.
this simplifies to:
x = (-7 +/- 63) / 8
your solutions are:
x = 7
x = -8.75
since you can't take the log of a negative number, the solution of x = -8.75 is not good.
the only solution left is x = 7.
you would need to confirm this solution is good by replacing x with 7 in the original equation to see if the equation is true.
the original equation is:
log3(x+2) + log(3(4x-1) = 5
replace x with 7 and simplify to get:
log3(9) + log3(27) = 5
log3(9) = y if and only if 3^y = 9
this makes y = 2, so log3(9) = 2
similarly, log3(27) = y if and only if 3^6 = 27
this makes y = 3, so log3(27) = 3
your equation of:
log3(9) + log3(27) = 5 becomes:
2 + 3 = 5 which is true.
this confirms the solution is good.
x = 7 is your solution.
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