Aptitude - Logarithm - Discussion
Discussion Forum : Logarithm - General Questions (Q.No. 6)
6.
| If log10 7 = a, then log10 |  |
1 |  |
is equal to: |
| 70 |
Answer: Option
Explanation:
|
= log10 1 - log10 70 | |||||
| = - log10 (7 x 10) | ||||||
| = - (log10 7 + log10 10) | ||||||
| = - (a + 1). |
Discussion:
33 comments Page 1 of 4.
Rohit said:
5 years ago
By simply log10(1/70) is explained as:
Log (a/b) formula.
Log a-logb
So,
Log10(1) -log10(70).
In this step log10(1) is 0.
So,
0- log10(70)
We write this step in another way as like.
-log10(7*10).
This step is in the form of;
Log (a*b) =log a+ logb.
So,
-[log10(7) +log10(10) ]
Log10(7) =a
And log10(10) =1.
- (a+1).
Log (a/b) formula.
Log a-logb
So,
Log10(1) -log10(70).
In this step log10(1) is 0.
So,
0- log10(70)
We write this step in another way as like.
-log10(7*10).
This step is in the form of;
Log (a*b) =log a+ logb.
So,
-[log10(7) +log10(10) ]
Log10(7) =a
And log10(10) =1.
- (a+1).
Nkayagwa Issa said:
9 years ago
Given,
Log7 = a,
Required to find = log10(1/70)
From logarithim of quotient,
Loga/b = loga - logb,
Therefore, log10(1/70) = log1 - log70.
= 0 - log70.
Since;
Any number to power one is 0 (logs laws).
= - log(10 + 7).
= -(1 + a).
Therefore, = -(1 + a). Because log7 is already given which is a and log10 = 1.
Log7 = a,
Required to find = log10(1/70)
From logarithim of quotient,
Loga/b = loga - logb,
Therefore, log10(1/70) = log1 - log70.
= 0 - log70.
Since;
Any number to power one is 0 (logs laws).
= - log(10 + 7).
= -(1 + a).
Therefore, = -(1 + a). Because log7 is already given which is a and log10 = 1.
Vinayak said:
1 decade ago
As we know that,
log 10 1 = 0.
And
log x x =1.
log 10 (1/70)=log 10 1 - log 10 70.
= 0 - log 10 70.
= - log 10 (7*10).
= -(log 1o 7 + log 10 10).
= -(a+1).
Because log10 7 = a is given and log x x =1.
log 10 1 = 0.
And
log x x =1.
log 10 (1/70)=log 10 1 - log 10 70.
= 0 - log 10 70.
= - log 10 (7*10).
= -(log 1o 7 + log 10 10).
= -(a+1).
Because log10 7 = a is given and log x x =1.
Tashi said:
1 decade ago
log 10 (1/7) = log10 -log10 70.
log10 1 = 0.
Therefore 0-log10 70.
-log10 70 = its same as -log10 (7*10).
-(log10 7+log10 10).
Given log10 7 = a.
And log10 10 equal to 1 = -(a+1).
Therefore Log10 (1/70) = -(a+1).
log10 1 = 0.
Therefore 0-log10 70.
-log10 70 = its same as -log10 (7*10).
-(log10 7+log10 10).
Given log10 7 = a.
And log10 10 equal to 1 = -(a+1).
Therefore Log10 (1/70) = -(a+1).
Sidhu said:
9 years ago
As we know that,
log10 1 = 0.
And
log x x =1.
log10 (1/70) = lo10 1 - log10 70.
= 0 - log10 70.
= - log10 (7 * 10).
= - (log10 7 + log10 10).
= - (a + 1).
Because log10 7 = a is given and log x x =1.
log10 1 = 0.
And
log x x =1.
log10 (1/70) = lo10 1 - log10 70.
= 0 - log10 70.
= - log10 (7 * 10).
= - (log10 7 + log10 10).
= - (a + 1).
Because log10 7 = a is given and log x x =1.
Prakash said:
10 years ago
log (1/70) = log (1/7*10).
= log (1/7*1/10).
= log (1/7)+log (1/10).
= -log 7-log 10.
= -a-1.
= -(a+1).
= log (1/7*1/10).
= log (1/7)+log (1/10).
= -log 7-log 10.
= -a-1.
= -(a+1).
Swathi said:
1 decade ago
(log m-log n) is log(m\n) so we get log(70\1) tat is log(70) which can be written as log(7*10)
Gosai devam said:
10 years ago
log 10(1) - log 10(70).
= 1-log 10(7*10).
-log 10(7)+log 10(10).
-a+1 answer.
= 1-log 10(7*10).
-log 10(7)+log 10(10).
-a+1 answer.
Akanthreddy said:
9 years ago
2 step = log 10 70 = we can write it has 10*7 = 70 so we get that 2 nd step.
Tanooja said:
7 years ago
log10(1/70) = -log10(70),
= -log10(7*10),
= -(log10(7) + log10(10)) =-(a+1).
= -log10(7*10),
= -(log10(7) + log10(10)) =-(a+1).
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