Before solving the equation, we'll have to impose constraints of existence of logarithms.

x>0

Now, we could apply the power property of logarithms:

3log 5 = log 5^3 = log 125

3log x = log x^3

We'll re-write the equation:

log x^3 + log 5^3 = log 6

Now, we'll apply the product property of logarithms:

log x^3 + log 5^3 = log x^3 * 5^3

log x^3 * 5^3 = log 6

Because the logarithms from both sides have matching bases, we'll use the one to one property:

x^3 * 5^3 = 6

x^3 = 6/5^3

x = (6/5^3)^1/3

**x = [(6)^1/3]/5 >0**

Because the solution is in the interval (0, +inf.), the solution is admissible.

3logx+3log5 = log6.

Solution:

By property of logarithms, m* log a = loga^m. and log a - lob = log(a/b).

So 3 log x = logx^3.

Therefore the given equation could be written as:

log x^3 +log5^3 = log6. Subtract log5^3.

log x^3 = log6 - log5^3

logx^3 = log (6/5^3). Take antilogarithms.

x^3 = 6/5^3. Take cube root.

x = 6^(1/3)/5 = 0.363424118 nearly.