Question 1204590
i get x = 3.537679573
rplace x with that in the original equation to get:
485 * 5^(x+2) = 12^(2x-1)
you will get 3600921.501 = 3600921.501, confirming the value of x is good.


here's how i did it.


start with 485 * 5^(x+2) = 12^(2x-1)
take the log of both sides of the equation to get:
log(485 * 5^(x+2)) = log(12^(2x-1))
this becomes:
log(485) + log(5^(x+2)) = log(12^(2x-1)), which becomes:
log(485) + (x+2) * log(5) = (2x-1) * log(12), which becomes:
log(485) + x * log(5) + 2 * log(5) = 2x * log(12) - log(12)
place all the terms with x on the right side of the equation and all the terms without x on the left side of the equation to get:
log(485) + 2 * log(5) + log(12) = 2x * log(12) - x * log(5)
factor out the x on the right sitde of the equation to get:
log(485) + 2 * log(5) + log(12) = x * (2 * log(12) - log(5))
divide both sides of the equation by (2 * log(12) - log(5) to get:
(log(485) + 2 * log(5) + log(12) / (2 *  log(12) - log(5)) = x
solve for x to get:
x = 3.537679573.


that's your answer.


the log and arithmetic rules that i used are:


log(a * b) = log(a) + log(b)
(a + b) * log(c) = a * log(c) + b * log(c)
log(a^b) = b * log(a)
x * log(a) + x * log(b) = x * (log(a) + log(b))
if a = b, then log(a) = log(b)


the a and b and x in the rules are not to be confused with the variable with the same names in the actual exectuion of the rules.
for example:


if a = b, then log(a) = log(b) is used in the statements:
start with 485 * 5^(x+2) = 12^(2x-1)
take the log of both sides of the equation to get:
log(485 * 5^(x+2)) = log(12^(2x-1))


(a + b) * log(c) = a * log(c) + b * log(c) is used in the statements:
log(485) + (x+2) * log(5) = (2x-1) * log(12), which becomes:
log(485) + x * log(5) + 2 * log(5) = 2x * log(12) - log(12)


a reference on log rules is shown below:
<a href = "https://sites.nicholas.duke.edu/statsreview/math/basic-math-logarithms/" target = "_blank">https://sites.nicholas.duke.edu/statsreview/math/basic-math-logarithms/</a>