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

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Answer 1 · 2016-02-20T09:07:58

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

#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))#