We have numerator = $8^x - 2^x = (2^3)^x - 2^x = (2^(3x) - 2^x = 2^x(2^{2x} - 1)= 2^x(2^x + 1)(2^x-1)$
Denominator = $6^x - 2^x = 3^x(2^x-1)$
So $\frac{8^x - 2^x }{6^x - 3^x} = \frac{2^x(2^x+1)}{3^x} = 2$
Or $2^x(2^x+1) = 2 * 3^x$
x cannot be zero as in the original question x = 0 is not possible
So we must ahve $2^x = 2$ and $2^x+ 1 = 3^x$ by equating even on both sides and odd on both sides
1st equuation gives x = 1 and it satisfies 2nd equation as well.
so Solution x = 1
No comments:
Post a Comment