Use the laws of exponents to solve for x in the equation 729^4x−1=93^x+8 .(1 point)
Responses
1.) x=11/9
2.) x= -3/2
3.) x=9
4.) 19/6
To solve for x in the equation 729^(4x-1) = 93^(x+8), we can start by taking the logarithm of both sides of the equation.
Using the laws of logarithms, we can rewrite the equation as:
(4x-1) log (729) = (x+8) log (93)
Next, we can simplify the logarithms using the properties of exponents:
(4x-1) log (3^6) = (x+8) log (93)
Now, substitute the value of log (3^6) = 6 log (3) and log (93) into the equation:
(4x-1) (6 log (3)) = (x+8) log (93)
Expand the equation:
24x log (3) - 6 log (3) = x log (93) + 8 log (93)
Combine like terms:
24x log (3) - x log (93) = 8 log (93) + 6 log (3)
Move the x terms to one side of the equation and the log terms to the other side:
24x log (3) - x log (93) - 8 log (93) = 6 log (3)
Factor out x from the x terms:
x (24 log (3) - log (93)) - 8 log (93) = 6 log (3)
Divide both sides of the equation by (24 log (3) - log (93)):
x = (6 log (3) + 8 log (93)) / (24 log (3) - log (93))
Simplify the expression:
x = (2 log (3) + 8 log (93)) / (8 log (3) - log (93))
Now, use the value of log (3) = 0.477 and log (93) = 1.968, and simplify the expression further:
x = (2 * 0.477 + 8 * 1.968) / (8 * 0.477 - 1.968)
x = (0.954 + 15.744) / (3.816 - 1.968)
x = 16.698 / 1.848
x ≈ 9.031
Therefore, the value of x is approximately 9.