log 2^(x-1)= log 5^(x-2)log2x−1=log5x−2
These logarithms have the same base
Log_a y = Log_a xlogay=logax
a ^(Log_a x) = a^(Log_a y) ---------1 alogax=alogay−−−−−−−−−1
Now a ^(log_a n) = nalogan=n
Extend this logic to 1
We get
x = yx=y
Now lets use this result
Log_a y = Log_a x iff x = ylogay=logax⇔x=y
So we get
2^(x-1) = 5^(x-2)2x−1=5x−2
2^(x-1) = 5^(x-1 - 1)2x−1=5x−1−1
2^(x-1) = (5^(x-1))/52x−1=5x−15
2^(x-1)/ (5^(x-1)) = 1/52x−15x−1=15
(2/5)^(x-1 ) = 1/5(25)x−1=15
0.4)^(x-1 ) = 0.20.4)x−1=0.2
x-1 = Log_(0.4) 0.2x−1=log0.40.2
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))