log(base5)x+log(base25)x+log(base125)x=33
solve for x
let's establish a property
let log(base25) x = y
25^y = x
5^(2y) = x
back to log notation ...
log(base5)x = 2y
y = (1/2)log(base5) x
so log(base5)x+log(base25)x+log(base125)x=33
log(base5)x + (1/2)log(base5)x + (1/3)log(base5)x = 33
(11/6)log(base5)x = 33
log(base5)x = 18
x = 5^18
log base100 power1
To solve the equation log(base5)x + log(base25)x + log(base125)x = 33, we can use the properties of logarithms. Specifically, we can use the property that states: log(baseA)(B) + log(baseA)(C) = log(baseA)(BC).
Let's rewrite the equation using this property:
log(base5)x + log(base25)x + log(base125)x = 33
Using the property, we can combine the logarithms:
log(base5)(x * x * x) = 33
Now, we have the equation:
log(base5)(x^3) = 33
To eliminate the logarithm, we can rewrite the equation in exponential form:
5^33 = x^3
Taking the cube root of both sides:
x = (5^33)^(1/3)
Simplifying further:
x = 5^(33/3)
Now, let's calculate and simplify the value for x:
x ≈ 5^(11)
Finally, calculating the value:
x ≈ 488,281,250
Therefore, the solution to the equation log(base5)x + log(base25)x + log(base125)x = 33 is approximately x ≈ 488,281,250.