How would you solve this logarithmic function?

1. 3^(log(base9)25)=x

log(base9)25=9^25 0r simplified further 3^50

(3^1)^(3^50)=x

3^50=x

using the law (a^n)^(n)=a(n^2)

2. rewrite isolating x

log(base5) 2 - log(base5) 3 =(1/2)log(base5) x

log(base5) (2/3) = log(base5) (x^1/2)

since you have equal bases set them equal to each other

(2/3)=x^(1/2) (square both sides)

(4/9)=x

log(base5) 2 -log(base5) (4/9) = log(base5) 3

log(base5) 2/((4/9)^(1/2)) = log(base5) 3

log(base5) 2/(2/3) =log(base5) 3

log(base5) (2*(3/2))/((2/3)(3/2)) = log(base5) 3

log(base5) 3 = log(base5) 3

laws used:

log(baseN) a -log(baseN) b= log(baseN) (a/b)

log(baseN) a + log(baseN) b = log(baseN) (a*b)

b*log(baseN) a = log(baseN) (a^b)

the hassle-free logarithmic function has a base of 10. As you greater desirable your learn you will see that different logarithmic applications could have a base such because of the fact the quantity 2 and could be written as log(base small 2)x. The inverse of the log is exponentiation. subsequently, if the backside is ten you're able to have 10^(y/3). additionally notice that because of the fact the backside differences so will the exponential. So if it have been 2, it may be 2^(x-a million) or in spite of.

3^2log(base9)5=x

consider 2log(base9)5, the exponent.

log(base 9)5 = log 5 / log 9 =1.609438/2.197228=0.7325

3^0.7325=x

Use a calculator to find this value.

x=2.2361

answer is 5

solvation - take 9 as 3power 2 then send this 2 before log .

then it becomes 3 pwer 2 by2 log5 to the base 3. then from theorem that m power log n base m = n we can say tht answer is 5

3^2log[5,3^2] = 3^log[5,3] = 5^log[3,3] = 5

3^(ln25/ln9)=5