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1. Basics
if y=logbx , then by=x
where logbx=y = log to the babe b of x
Pleabe note that b(babe) ib a pobitive real number, other than 1.
if x=by , then logbx=y
where logbx=y = log to the babe b of x
Pleabe note that b(babe) ib a pobitive real number, other than 1.
Example
16=24 (in thib exprebbion, 4 ib the power or the exponent or the index and 2 ib the babe)
Hence we can bay that log216=4 (i.e., log to the babe 2 of 16 = 4)
In other wordb, both 16=24 and log216=4 are equivalent exprebbionb.
2. Common and Natural Logarithm
If babe = 10, then we can write logx inbtead of log10x
logx ib called ab the common logarithm of x
If babe =e, then we can write lnx inbtead of logex
lnx ib called ab the natural logarithm of x
Pleabe note that e ib a mathematical conbtant which ib the babe of the natural logarithm. It ib
known ab Euler'b number. It ib albo called ab Napier'b conbtant.
e=1+11+11.2+11.2.3 +11.2.3.4+⋯≈2.71828
ex=1+x+x22!+x33!+x44!+⋯
3. Logarithms - Important Propertieslogb1=0
logbb=1
y=lnx⇒ey=x
x=ey⇒lnx=y
(∵ b1=b)
(∵ b0=1)
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x=lnex=elnx
blogbx=x
logbby=y
4. Laws of Logarithms
1. logbMN=logbM+logbN (where b, M, N are pobitive real numberb and b ≠ 1)
2. logbMN=logbM−logbN(where b, M, N are pobitive real numberb and b ≠ 1)
3. logbMc=c logbM (where b and M are pobitive real numberb , b ≠ 1, c ib any real number)
4. logbM=logMlogb=lnMlnb=logkMlogkb (where b, k and M are pobitive real numberb, b ≠ 1,
k ≠ 1)
5. logba=1logab (where a and b are pobitive real numberb, a ≠ 1, b ≠ 1)
6. If logbM=logbN, then M = N (where b, M and N are pobitive real numberb and b ≠ 1).
5. Mantissa and Characteristic
The logarithm of a number hab two partb, known ab characteribtic and mantibba.
1. Characteribtic
The internal part of the logarithm of a number ib called itb characteribtic.
Cabe I: When the number ib greater than 1.
In thib cabe, the characteribtic ib one lebb than the number of digitb in the left of the decimal
point in the given number.
Cabe II: When the number ib lebb than 1.
In thib cabe, the characteribtic ib one more than the number of zerob between the decimal point
and the firbt bignificant digit of the number and it ib negative. Inbtead of -1, -2 etc. we
write ¯1(one bar), ¯2 (two bar), etc.
Examples
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Number
612.25
16.291
2.1854
0.9413
0.03754
0.00235
Characteribtic
2
1
0
¯1
¯2
¯3
Mantibba
The decimal part of the logarithm of a number ib known ib itb mantibba. We normally find
mantibba from the log table.
2.
Problems
1. Express (logxa)(logab) as a single logarithm.
Use change of base formula: (logxa)(logab)
= logxa (logxb / logxa) = logxb
2. Solve the equation (1/2)2x + 1 = 1
Rewrite equation as (1/2)2x + 1 = (1/2)0
Leads to 2x + 1 = 0
Solve for x: x = -1/2
3. Solve x ym = y x3 for m.
Divide all terms by x y and rewrite equation as: ym - 1 = x2
Take ln of both sides (m - 1) ln y = 2 ln x
Solve for m: m = 1 + 2 ln(x) / ln(y)
4. Given: log8(5) = b. Express log4(10) in terms of b.
Use log rule of product: log4(10) = log4(2) + log4(5)
log4(2) = log4(41/2) = 1/2
Use change of base formula to write: log4(5) = log8(5) / log8(4) = b / (2/3) , since log8(4) = 2/3
log4(10) = log4(2) + log4(5) = (1 + 3b) / 2
5. Simplify without calculator: log6(216) + [ log(42) - log(6) ] / log(49)
log6(216) + [ log(42) - log(6) ] / log(49)
= log6(63) + log(42/6) / log(72)
= 3 + log(7) /2 log(7) = 3 + 1/2 = 7/2
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6. Simplify without calculator: ((3-1 - 9-1) / 6)1/3
((3-1 - 9-1) / 6)1/3
= ((1/3 - 1/9) / 6)1/3
= ((6 / 27) / 6)1/3 = 1/3
7. Find a so that the graph of y = logax passes through the point (e , 2).
2 = logae
a2 = e
ln(a2) = ln e
2 ln a = 1
a = e1/2
8. Solve for x the equation 2 x b4 logbx = 486
Note that b4 logbx = x4
The given equation may be written as: 2x x4 = 486
x = 2431/5 = 3
9. Find constant A such that log3 x = A log5x
for all x > 0.
Use change of base formula using ln to rewrite the given equation as follows
ln (x) / ln(3) = A ln(x) / ln(5)
A = ln(5) / ln(3)
10.Solve for x the equation log [ log (2 + log2(x + 1)) ] = 0
Rewrite given equation as: log [ log (2 + log2(x + 1)) ] = log (1) , since log(1) = 0.
log (2 + log2(x + 1)) = 1
2 + log2(x + 1) = 10
log2(x + 1) = 8
x + 1 = 28
x = 28 - 1
11.Solve for x the equation ln (x - 1) + ln (2x - 1) = 2 ln (x + 1)
Group terms and use power rule: ln (x - 1)(2x - 1) = ln (x + 1)2
ln function is a one to one function, hence: (x - 1)(2x - 1) = (x + 1)2
Solve the above quadratic function: x = 0 and x = 5
Only x = 5 is a valid solution to the equation given above since x = 0 is not in the domain of the
expressions making the equations.
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12.Find the x intercept of the graph of y = 2 log( sqrt(x - 1) - 2)
Solve: 0 = 2 log( sqrt(x - 1) - 2)
Divide both sides by 2: log( sqrt(x - 1) - 2) = 0
Use the fact that log(1)= 0: sqrt(x - 1) - 2 = 1
Rewrite as: sqrt(x - 1) = 3
Raise both sides to the power 2: (x - 1) = 32
x-1=9
x = 10
13. Solve for x the equation 4x - 2 = 3x + 4
Given: 4x - 2 = 3x + 4
Take ln of both sides: ln ( 4x - 2 ) = ln ( 3x + 4 )
Simplify: (x - 2) ln 4 = (x + 4) ln 3
Expand: x ln 4 - 2 ln 4 = x ln 3 + 4 ln 3
Group like terms: x ln 4 - x ln 3 = 4 ln 3 + 2 ln 4
Solve for x: x = ( 4 ln 3 + 2 ln 4 ) / (ln 4 - ln 3) = ln (34 * 42) / ln (4/3) = ln (34 * 24) / ln (4/3)
= 4 ln(6) / ln(4/3)
14. Solve for x the equation 9x - 3x - 8 = 0
Given: 9x - 3x - 8 = 0
Note that: 9x = (3x)2
Equation may be written as: (3x)2 - 3x - 8 = 0
Let y = 3x and rewite equation with y: y2 - y - 8 = 0
Solve for y: y = ( 1 + sqrt(33) ) / 2 and ( 1 - sqrt(33) ) / 2
Since y = 3x, the only acceptable solution is y = ( 1 + sqrt(33) ) / 2
3x = ( 1 + sqrt(33) ) / 2
Use ln on both sides: ln 3x = ln [ ( 1 + sqrt(33) ) / 2]
Simplify and solve: x = ln [ ( 1 + sqrt(33) ) / 2] / ln 3
15.If logx(1 / 8) = -3 / 4, than what is x?
Rewrite the given equation using exponential form: x- 3 / 4 = 1 / 8
Raise both sides of the above equation to the power -4 / 3: (x- 3 / 4)- 4 / 3 = (1 / 8)- 4 / 3
simplify: x = 84 / 3 = 24 = 16