Question 1200017
a) 3^x=5(2^x)


divide both sides of the equation by 2^x to get:
3^x/2^x = 5
this becomes:
(3/2)^x = 5
take the log of both sides of the equation to getL
log((3/2)^x) = log(5)
this becomes:
x * log(3/2) = log(5)
solve for x to get:
x = log(5)/log(3/2)  = 3.969362296.
that's our solution.
confirm by replacing x with that in the original equation to get:
3^x = 5 * 2^x becomes 78.31899737 = 78.31899737, confirming the the solution is good.


b) 9^(x-1)*3^x=27


9^(x-1) is equal to 3^2)^(x-1) which is equal to 3^(2*(x-1) which is equal to 3^(2x-2)
your equation becomes:
3^(2x-2) * 3^x = 27
this becomes 3^(2x-2+x) = 27 which becomes 3^(3x-2) = 27
take the log of both sides of the equation to get:
log(3^(3x-2)) = log(27) which becomes (3x-2)*log(3) = log(27)
divide both sides of the equation by log(3) to get:
3x-2 = log(27)/log(3)
add 2 to both sides of the equation and then divide both sides of the equation by 3 to get:
x = (log(27)/log(3)+2)/3
solve for x to get:
x = 1.666666666..... which is equal to 1 + 2/3 which is equal to 5/3.
replace x with that in the original equation to get:
9^(x-1)*3^x=27 becomes 9^(5/3 - 1) * 3^(5/3) = 27 which becomes 27 = 27, confirming the soluton is correct.
your solution is that x = 5/3 or x = 1.6666667.
5/3 is the exact solution whiile x = 1.6666667 is an approximation rounded to the number of decimal places that can be displayed by the calculator.


use of the following properties have been made of.
log(b^x) = x * log(b)
a^x * a^y = a^(x + y)
(a^2)^y = a^(2y)