You may also use the properties of exponentials, such that:

`a^x*b^x = (a*b)^x`

Reasoning by analogy yields:

`(5^(3x))*(7^(2x)) = (5^3)^x*(7^2)^x`

`(5^(3x))*(7^(2x)) = (5^3*7^2)^x => (5^(3x))*(7^(2x) = 6125^x`

Since `(5^(3x))*(7^(2x)) = 3` yields:

`6125^x = 3 => ln (6125^x) = ln 3 => x = ln 3/ln 6125`

`x = 1.098/8.720 => x = 0.125`

**Hence, evaluating the solution to the given equation yields **`x = 0.125.`

We'll take logarithms both sides:

ln [(5^3x)*(7^2x)] = ln 3

We'll apply the product rule of logarithms:

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

ln [(5^3x)*(7^2x)] = ln (5^3x) + ln (7^2x)

We'll apply power rule of logarithms:

ln (5^3x) = 3x*ln 5

ln (7^2x) = 2x*ln 7

3x*ln 5 + 2x*ln 7 = ln 3

We'll factorize by x:

x*(3*ln 5 + 2*ln 7) = ln 3

x = ln 3/(3*ln 5 + 2*ln 7)

x = ln 3/(ln 125*49)

x = ln 3/ ln 6125