You certainly took the long route to get the same answer. Factoring the bottom gives you a denominator that cancels one of the numbers in the numerator. So much faster.
folks are saying that you took the long wday. Here is my solution - 5^3x /( 3*5^x ) = 5 5^2x / 3 = 5 5^x /3 = 1 5^x = 3 x log 5 = log 3 x = log3/log5 = .688606
Great solution !
You certainly took the long route to get the same answer. Factoring the bottom gives you a denominator that cancels one of the numbers in the numerator. So much faster.
Thanks.
Very good, thanks.
Nice
Долго, на 3:00 логарифмировать по базе 5
Получаем 2х=log(5)15
x=(log(5)15)/2
Зачем это все нужно дальше?
👍
You took the long way, there is a much faster way
You show a long way to solve this equation for x.
x = (log 15)/(log 25).
folks are saying that you took the long wday.
Here is my solution -
5^3x /( 3*5^x ) = 5
5^2x / 3 = 5
5^x /3 = 1
5^x = 3
x log 5 = log 3
x = log3/log5 = .688606
(5^3×)/3(5^×)
=(1/3)(5^2×)
5^2×=3(5)
=> (×)=(1/2){log3+log5}/(log5)
=(1/2){1+log @3 (5)}
5^x = a (a･a･a)/(a+a+a) = 5 a³/3a = 5 a²/3 = 5 a² = 15
a = 5^x = √15 x = log₅√15
4^2.5 or 4^2.25 ?
Nice