Aptitude - Logarithm - Discussion

Discussion :: Logarithm - General Questions (Q.No.2)

2. 

If log 2 = 0.3010 and log 3 = 0.4771, the value of log5 512 is:

[A]. 2.870
[B]. 2.967
[C]. 3.876
[D]. 3.912

Answer: Option C

Explanation:

log5 512
= log 512
log 5
= log 29
log (10/2)
= 9 log 2
log 10 - log 2
= (9 x 0.3010)
1 - 0.3010
= 2.709
0.699
= 2709
699
= 3.876

Happy said: (Apr 23, 2011)  
How is log 10=1.

Sumit said: (May 27, 2011)  
How can you get log (10/2) ?

Kasthuri said: (Aug 6, 2011)  
If log 2 = 0.3010 and log 3 = 0.4771,and log 7 = 0.8451 then the value of log75 is?

Jithu said: (Aug 19, 2011)  
@kasthuri. Can you explain the solution ?

Sandy said: (Aug 20, 2011)  
log 10 means
log 10 to the base 10
that ts equal to 1
so,
log10=1

Heaven.Bangladesh said: (Sep 1, 2011)  
@sumit

Log 5=log(10/2).coz,10/2=5

Shalini said: (Sep 3, 2011)  
@Kasthuri

log 75= log(5*5*3) = log(5*5)+log(3), because log m*n=log m+log n
=2 log5 +log3
=2 log(10/2)+log3
=2(log10-log2)+log3
=2(1-.3010)+..4771
=2*.699+.4771
=1.398+.4771
=1.8751

Sinkay said: (Dec 6, 2011)  
If log 2=0.3010 and log 7=0.8451,

Usha said: (Feb 22, 2013)  
If log2=0.3010 and log3= 0.4771, then log75 = ?

Jeromy said: (Mar 15, 2013)  
Why are they dividing log 512 by log 5?

Vigneshraj said: (Mar 25, 2013)  
@jeromy.

Click the impotant formulas link there you can see 7th formula using that formula we divide log512 by log5.

Lekan said: (Nov 1, 2013)  
Please who can solve this log10 2 = 0.3010?

K.Rajeshreddy said: (Jun 1, 2014)  
log5512 = log512/log5 = 9log2/log5.
= 9log2/log3+log2.
= 9*0.3010/0.4771+0.3010.

I think this is this the correct procedure as log 2 and log3 values are given. Why we need to go for log 10/2?

Apoorva said: (Jul 6, 2014)  
Why is it only 2?

Nirav said: (Jul 18, 2014)  
Why are they dividing log 512 by log 5?

Is this a formula?

Nagdish Kewat said: (Aug 19, 2014)  
What is the answer of log3(log2(log3t)) =1 t=?

Raj Kumar said: (Aug 27, 2014)  
@Nirav.

log512/log5 = 9log2/log(2+3) = 9log2/(log2*log3).

= 9*.3010/(.3010*.4771).

= 9/.4771 >18.

Saurabh said: (Aug 30, 2014)  
Why are they dividing log 512 by log 5?

Vipin Mishra said: (Sep 5, 2014)  
Formula:

log a x = log x/log a.

log 5 512 = log 512/log 5.

Goose said: (Sep 19, 2014)  
log(75) = log(70+5).

= log(7*10)+(2+3).

= [log(7)+log(10)]*log(2)*log(3).

= [0.8541+1]*0.3010*0.4771.

= 1.8451*0.3010*0.4771.

= 0.1557.

Dhivi said: (Feb 14, 2015)  
Can you say the answer for the (5/512) power 1/9?

Vaishu said: (Jun 10, 2015)  
If log 2 10 = 0.3010 and log 7 10 = 0.8451, then the value of log 2.8?

Sumeet said: (Jun 28, 2015)  
{log(10/2) = log5} Can we write like this? I have a doubt!

Sumeet said: (Jun 28, 2015)  
If log(10/2) = log5, then log(512/5) should be equal to log(9/1). Isn't it?

Kate said: (Aug 14, 2015)  
Please I don't know anything about math I need help forms you guys.

Ajitha said: (Nov 3, 2015)  
Can you explain it once for me?

Ajitha said: (Nov 3, 2015)  
log5 512 = log 512/log5 by which formula.

Ankur said: (Nov 23, 2015)  
Answer is right, but too long, tell in short.

Girish said: (Dec 9, 2015)  
That is base change rule it seems log (b) to the base a, = log (b) to base c/log (a) to base c i.e log 512/log5.

Pavan S@I said: (Mar 29, 2016)  
How to write log 4 power 2013, if log 2 = 0.3010? Please help me.

Rita said: (Apr 3, 2016)  
Find the value of log2 + log3?

Aman Srivastava said: (Apr 16, 2016)  
Which property of log is used in:- log5 512 = log 512/log 5.

Grace Lancelot said: (Jun 21, 2016)  
From where did log10 - log2 came from?

Koala said: (Jul 4, 2016)  
It is a basic formula of logarithm.

If,

log(a/b) = log a - log b

log (a.b) = log a + log b

Akash said: (Jul 17, 2016)  
If log (a + b)/7 = loga + logb/2 then show that a/b + b/a = 47. Can anyone solve this problem?

Shyam said: (Aug 9, 2016)  
It's quite confusing.

Rakesh.H said: (Aug 17, 2016)  
Simple way;

In 512/ln 5 = 3.876.

Naveen said: (Aug 28, 2016)  
If log 3 = .4771 than log 27 = ?

Sarah said: (Sep 3, 2016)  
How did you know log 512 = 2^9?

Kennedy said: (Sep 17, 2016)  
Please explain the last part are you supposed to divide or subtract?

Sukh Singh said: (Oct 4, 2016)  
How to use Log3?

Please explain me.

Arban said: (Oct 22, 2016)  
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?

Michael Essandoh said: (Nov 3, 2016)  
Please solve this, given that log 9 = 0.9542 and log 6 = 0.7781, find the value of log base 10 of 225.

Saffee2 said: (Nov 30, 2016)  
@Michael Essandoh.

Is it 2.5104?

Anu said: (Jan 4, 2017)  
How to subtract 1 - 0.3010 =? Explain.

Shreyash said: (Jan 5, 2017)  
@Arban.

Whenever base is not mentioned, it is taken as 10.
So, log 10 =1 (as the base here is 10).

Mansi said: (Feb 21, 2017)  
Here Log2 =0.3010.
Then, Find log 5.

Mushariq said: (Mar 2, 2017)  
Nice explanation, thank you for all your solution.

Ayesha said: (Jun 22, 2017)  
If log2=0.3010 then find the value of log32. Can someone solve this?

Siddharth said: (Jun 29, 2017)  
@Ayesha.

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

Manasa said: (Jul 21, 2017)  
Superb explanation, Thank you all.

Dheeraj said: (Jul 22, 2017)  
@Naveen.

log27=log3^3=3log3=3 * 0.4771=1.4313.

Ankush Premi said: (Aug 1, 2017)  
How to solve this question 1-log5= 1/3 ( log1/2+logx +1/3 log5)?

Can anyone solve this

Bittu said: (Oct 5, 2017)  
Why do we take log base 10 in log table?

Angel said: (Dec 18, 2017)  
Given that log 2 =0.3010 and log 3=0.4771 what is log 1.125?

Dinesh Puri said: (Dec 29, 2017)  
When log3=0.4771 and log 5=0.690then find value of log 4=?

Please solve this.

Sweta said: (May 28, 2018)  
@Dinesh Puri.

log 4=log 2^2.
=2log2,
=2log(10/5),
= 2*{log10-log5},
= 2*{1-.690},
= 0.62.

Aman Mittal said: (Jun 6, 2018)  
Log base is 5 and exponential is 512 so how you kept log5=log512?

Simran said: (Sep 4, 2018)  
Can anyone explain it properly?

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