Aptitude - Logarithm - Discussion

Logarithm base change rule
The base b logarithm of x is base c logarithm of x divided by the base c logarithm of b.

logb(x) = logc(x) / logc(b)

For example, in order to calculate log2(8) in calculator, we need to change the base to 10:

log2(8) = log10(8) / log10(2)

So by this we get
log5 512 = log 512 / log 5 with base 10

Now since 5 can be written as 10/2
so log 5 = log(10/2)

Now using property log(x / y) = log(x) - log(y)
so, log(10/2) can be written as log(10) - log(2)

So finally we got
log5 512 = log(512) / log(10) - log(2)
since 512 = 2^9

log 512 = log(2^9) = 9 log 2

= 9 log 2 / log 10 - log 2
= (9 x 0.3010) / 1 - 0.3010
= 2.709 / 0.699 = 3.876

Hope it helps !

No @Amrita.

Log 5 can't be written as log 2+log3.
But it can be written as log 5+ log 1.
Hope it helps.

Anyone, please Explain me in detail.

@Ayesha.

log32 = log(2^5)= 5log2=5*0.3010= 1.505.

Given that log 2 =0.3010 and log 3=0.4771 what is log 1.125?

Log base is 5 and exponential is 512 so how you kept log5=log512?

Can log 5 be written as log 2 + log 3?

Please explain the last part are you supposed to divide or subtract?

How to use Log3?

Please explain me.

How can log 10 be 1? It can only be 1 if the base is 10. If I remember the formula correctly then the base can be anything and is generally a random constant term (a, b etc. ). If we follow the same logic then the answer is most definitely wrong as per my calculations.

CAN ANYONE EXPLAIN THE CORRECT PROCESS?