Question

How do you solve `log 2^(x-1)= log 5^(x-2)`?

Answer 1

x = `(log(2)-2 log(5))/(log(2)-log(5) ) approx 2.75647`

Explanation:

`log 2^(x-1)= log 5^(x-2)`

These logarithms have the same base

`Log_a y = Log_a x`

`a ^(Log_a x) = a^(Log_a y) ---------1`

Now `a ^(log_a n) = n`

Extend this logic to 1

We get

`x = y`

Now lets use this result

`Log_a y = Log_a x iff x = y`

So we get

`2^(x-1) = 5^(x-2)`

`2^(x-1) = 5^(x-1 - 1)`

`2^(x-1) = (5^(x-1))/5`

`2^(x-1)/ (5^(x-1)) = 1/5`

`(2/5)^(x-1 ) = 1/5`

`0.4)^(x-1 ) = 0.2`

`x-1 = Log_(0.4) 0.2`

`x-1 = 1.75647...`

`x = 1.75647... + 1`

`x = 2.75647...`

If you need it perfect logs

x = `(log(2)-2 log(5))/(log(2)-log(5))`