Mental Logs Flashcards
How do you “undo” a log function?
Also known as the antilog function
When, log(x) = y
use x = 10y to undo the logarithm
All logarithm problems on the MCAT are base 10
Fundamentally, you are undoing the log function:
(your starting function): log(x) = y
(Raise both sides by the log’s base): 10log(x) = 10y
(Becuase 10log(x) is simply x): x = 10y
What is the equation for a basic log function?
log10(x) = y
All logarithm problems on the MCAT are base 10
What is log(10)=?
1
We can verify this by performing the antilog of our answer:
101 = 10
What is log(100)=?
2
We can verify this by performing the antilog of our answer:
102 = 100
What is log(1000)=?
3
We can verify this by performing the antilog of our answer:
103 = 1000
What is log(1/10)=?
-1
We can verify this by performing the antilog of our answer:
10-1 = 1/10 (becuase negative exponents place the basex in the denominator so that base-x = 1/basex
What is log(1/100)=?
-2
We can verify this by performing the antilog of our answer:
10-2 = 1/100 (becuase negative exponents place the 10x in the denominator so that 10-x = 1/10x
What is log(1/1000)=?
-3
We can verify this by performing the antilog of our answer:
10-3 = 1/10000 (becuase negative exponents place the basex in the denominator so that base-x = 1/basex
What is log(0.1/1)=?
-1, this is something you might see in a weak acid/base problem. However, do not let the fraction confuse you, this is simply log(1/10) but not in a reduced form.
What is the general rule for log(x) where x<1?
log(x) when x<1 will always result in a negative value
Write the expanded form of log(x/y)
log(x/y)=log(x)-log(y)
What is log(2)=?
Memorization Required
log(2) is an irrational number, but we estimate it to be ~0.3 to use on the MCAT
What is log(3)=?
log(3) is an irrational number, but we estimate it to be ~0.5 to use on the MCAT
What is log(5)=?
log(5) is also an irrational number, but we estimate it to be ~0.7 for mental math on the MCAT
What is log(0.2)=?
Will have to use memorized logs and expanded form to solve mentally
log(0.2) =-0.69897
Becuase log(0.2) =log(1/5)
we can expand this log, so that: log(1/5) = log(1) - log(5)
Since we memorized log(5), we can solve from here:
log(1)-log(0.5) = 0 - 0.7
Becuase log10(1) = 0
confirm by performing antilog:
100 = 1