Question

2^(x+1) = 3^(x-1)
could someone explain in detail how to solve these type of questions please?

Answer 1

Become friendly with logarithms. Because
\[\log(a^b) = b\cdot \log(a)\]
they are very useful in situations like this. For instance, taking the log of both sides,
\[\log(2^{x+1}) = (x+1)\log(2) = \log(3^{x-1}) = (x-1)\log(3)\]
Pulling the x's to one side and the constants to the other, we find
\[\left(\log(2)-\log(3)\right)x = -\left(\log(3)+\log(2)\right) \]
so we get
\[ x= \frac{\log(3)+\log(2)}{\log(3)-\log(2)} \approx 4.42 \]

Answer 2

As a side note, it doesn't actually matter which type of logarithm you use. Base 10, base e, base 2, it all works equally well.

Answer 3

how did you get log(2)-log(3))x= - (log(3)+log(2)) ?

Answer 4

Just rearrange the equation. Get all the x-terms on one side, and the constants on the other.

Answer 5

so first its like (x+1)log2= (x-1)log 3, right?

Answer 6

right... then expand that, to get
\[x\log(2) + \log(2) = x\log(3) - \log(3)\]

Answer 7

thanks !