Question

3log2 + 5logx -log256 = 1 find x

Answer 1

Question:

If 3log2 + 5logx - log256 = 1, then find the value of x.

Note:

• log(a) + log(b) = log(a•b)
• log(a) - log(b) = log(a/b)
• b•log(a) = log(a^b)

Solution:

We have;

=> 3log2 + 5logx - log256 = 1

=> log(2^3) + log(x^5) - log(2^8) = log2

{ note : considering base as 2 }

=> log{ (2^3)•(x^5)/2^8 } = log2

=> log{ x^5/2^5} = log2

=> x^5/2^5 = 2

=> x^5 = 2•2^5

=> x^5 = 2^6

=> x = 2^(6/5)

Or

=> x = 2•2^(1/5)

Note:

Considering different bases will provide different values of x.

Here,

In general, for this question we have;

x = 2•(base)^(1/5)