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.

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?

@Ayesha.

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

Simple way;

In 512/ln 5 = 3.876.

If log2=0.3010 then find the value of log32. Can someone solve this?

Nice explanation, thank you for all your solution.