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

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? |

Srini said: (Nov 11, 2019) | |

How did you get 0.3010? please, do explain. |

Amrita said: (Dec 18, 2019) | |

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

Shakti said: (Jan 2, 2020) | |

Sir is there any methods to solve 2709/699. It is complicated to solve soo. |

Harshini said: (Feb 21, 2020) | |

Why should we take log 5 only? |

Stella said: (Apr 27, 2020) | |

Hi, I am confused about the second step please help me and explain more. |

Habibur Rahman said: (Jun 20, 2020) | |

Anyone explain it easily. |

Sriya said: (Jul 11, 2020) | |

I am not getting this, Can anyone explain in correct way? |

Devilyne Ras said: (Dec 5, 2020) | |

Please try to explain in statement. |

Priya said: (Dec 10, 2020) | |

Please anyone explain. Please. |

Rohit Saini said: (Jan 21, 2021) | |

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 ! |

Kiprop said: (Jun 9, 2022) | |

Anyone, please Explain me in detail. |

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