How do you solve log 2^(x-1)= log 5^(x-2)log2x1=log5x2?

1 Answer
Feb 20, 2016

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

Explanation:

log 2^(x-1)= log 5^(x-2)log2x1=log5x2

These logarithms have the same base

Log_a y = Log_a xlogay=logax

a ^(Log_a x) = a^(Log_a y) ---------1 alogax=alogay1

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=logaxx=y

So we get

2^(x-1) = 5^(x-2)2x1=5x2

2^(x-1) = 5^(x-1 - 1)2x1=5x11

2^(x-1) = (5^(x-1))/52x1=5x15

2^(x-1)/ (5^(x-1)) = 1/52x15x1=15

(2/5)^(x-1 ) = 1/5(25)x1=15

0.4)^(x-1 ) = 0.20.4)x1=0.2

x-1 = Log_(0.4) 0.2x1=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))