# LOGARITHM WORKSHEET

1. Find the logarithm of 32 to the base 2.

2. Find the value of log2√31728.

3. Find the value of log(0.0001) to the base 0.1.

4. Find the value of log (1/81) to the base 9.

5. Find the value of log(0.0625) to the base 2.

6. Find the value of log(0.3) to the base 9.

7. If loga(√2) = 1/6, find the value of a.

8. Simplify : (1/2)log1025 - 2log103 + log1018.

9. Given log2 = 0.3010 and log3 = 0.4771, find the value of log6.

10. If 2logx = 4log3,  then find the value of x.

11. If logabc = x, logbca = y and logcab = z, then find the value of

1/(x + 1) + 1/(y + 1) + 1/(z + 1)

12. If a = log2412, b = log3624 and c = log4836, then find the value of (1 + abc) in terms of b and c.

13. If logx + logy = log(x + y), solve for y in terms of x. log232 = log2(2)5

= 5log2(2)

= 5(1)

= 5

Write 1728 as a power of 23.

1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

1728 = 26 x 33

1728 = 26 x [(3)2]3

1728 = 26 x (3)6

1728 = (23)6

log2√3(1728) = log2√3(23)6

Using the power rule of logarithms,

= 6log2√3(23)

= 6(1)

= 6

log0.10.0001 = log0.1(0.1)4

= 4log0.10.1

= 4(1)

= 4

log9(1/81) = log91 - log981

= 0 - log9(9)2

= -2log99

= -2(1)

= -2

log2(0.0625) = log2(0.5)4

= 4log2(0.5)

= 4log2(1/2)

= 4(log21 - log22)

= 4(0 - 1)

= 4(-1)

= -4

log9(0.3) = log9(1/3)

= log91 - log93

= 0 - log93

= -log93

= -1 / log39

= -1 / log332

= -1 / 2log33

= -1 / 2(1)

= -1/2

loga(√2) = 1/6

Write the equation in exponential form.

√2 = a1/6

Raise to the power 6 on both sides.

(√2)6 = (a1/6)6

(21/2)6 = a

23a

8 = a

= (1/2)log1025 - 2log103 + log1018

Using power rule of logarithms,

= log10251/2 - log1032 + log1018

= log10(52)1/2 - log1032 + log1018

= log105 - log109 + log1018

= log105 + log1018 - log109

Using the product rule of logarithms,

= log10(5 x 18) - log109

= log1090 - log109

Using the quotient rule of logarithms,

= log10(90/9)

= log1010

= 1

log6 = log(2 ⋅ 3)

= log2 + log3

Substitute the values of log2 and log3.

= 0.3010 + 0.4771

= 0.7781

2logx = 4log3

Divide each side by 2.

logx = 2log3

logx = log32

logx = log9

x = 9

x + 1 = logabc + logaa = logaabc

y + 1 = logbca + logbc = logbabc

z + 1 = logcab + logcc = logcabc

1/(x + 1) = 1/logaabc = logabca

1/(y + 1) = 1/logbabc = logabcb

1/(z + 1) = 1/logcabc = logabcc

1/(x + 1) + 1/(y + 1) + 1/(z + 1) = logabca + logabcb + logabcc

= logabcabc

= 1

1 + abc = 1 + log2412  log3624 ⋅ log4836

= 1 + log3612 ⋅ log4836

= 1 + log4812

= log4848 + log4812

= log48(48 ⋅ 12)

= log48(2 ⋅ 12)2

= 2log4824

= 2log3624 ⋅ log4836

= 2bc

logx + logy = log(x + y)

Using the product rule of logarithms on the left side of the equation,

log(xy) = log(x + y)

xy = x + y

Subtract y from both sides.

xy - y = x

Factor.

y(x - 1) = x

Divide both sides by (x - 1).

y = x/(x - 1)

## Video Lessons

Introduction to Logarithms

Fundamental Laws of Logarithms

Change of Base

Using Log Table

Using Antilog Table

Important Stuff

Difference between Bar Value and Negative Value

Multiplication of Two Logarithms

Relationship between Exponents and Logarithms

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Problem - 22

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