"Operations of Logarithms" Exponential function and logarithmic function PPT

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"Operations of Logarithms" Exponential function and logarithmic function PPT

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"Operations of Logarithms" Exponential function and logarithmic function PPT

Part One: Explanation of Curriculum Standards

1. Master the operational properties of logarithms, and be able to use them to simplify and evaluate.

2. Understand the base changing formula of logarithms and the application of its deformations.

3. Preliminarily master the application of logarithms in life.

Logarithmic operation PPT, part 2 content: independent preview

1. Operational properties of logarithms

1. (1) What are the algorithms for exponential calculation?

Tips: ①aras=ar+s(a>0,r,s∈Q);

②a^r/a^s =ar-s(a>0,r,s∈Q);

③(ar)s=ars(a>0,r,s∈Q);

④(ab)r=arbr(a>0,b>0,r∈Q).

(2) Calculate the values ​​of log24, log28 and log232. Can you analyze the operational relationship between the three?

Tip:∵log24=2,log28=3,log232=5,

∴log24+log28=log2(4×8)=log232;

log232-log28=log232/8=log24;

log232-log24=log232/4=log28.

(3) Calculate the values ​​of lg 10, lg 100, lg 1 000 and lg 104. What patterns can you find?

Tip: lg 10=1, lg 100=lg 102=2, lg 1 000=lg 103=3, lg 104=4, it can be seen that lg 10n=nlg 10=n.

2. Fill in the form

Operational properties of logarithms

3. Do it

(1) The result of simplifying 2lg 5+lg 4- 5^(log_5 2) is ()

A.0 B.2 C.4 D.6

Analysis: Original formula=2lg 5+2lg 2-2=2(lg 5+lg 2)-2=0.

Answer:A

(2) Judgment of right or wrong:

log3[(-4)×(-5)]=log3(-4)+log3(-5). ()

Answer: ×

Logarithmic operation PPT, the third part: inquiry learning

Application of properties of logarithmic operations

Example 1 Calculate the values ​​of the following equations:

(1)log2√(7/96)+log224-1/2log284;

(2)lg 52+2/3lg 8+lg 5·lg 20+(lg 2)2.

Analysis: Use the operational properties of logarithms to perform calculations.

Solution: (1) (Method 1) Original formula =log2(√7×24)/(√96×√84)=log21/√2=-1/2.

(Method Two)

Original formula=1/2log27/96+log2(23×3)-1/2log2(22×3×7)

=1/2log27-1/2log2(25×3)+3+log23-1-1/2log23-1/2log27

=-1/2×5-1/2log23+2+1/2log23=-5/2+2=-1/2.

(2)Original formula=2lg 5+2lg 2+lg 5×(1+lg 2)+(lg 2)2

=2(lg 5+lg 2)+lg 5+lg 2(lg 5+lg 2)

=2+lg 5+lg 2=2+1=3.

Reflect on the commonly used methods for simplification and evaluation of logarithms with the same base

(1) "Collect", take the sum (difference) of two logarithms with the same base into the logarithm of the product (quotient);

(2) "Split", split the logarithm of the product (quotient) into the sum (difference) of the logarithms.

The simplification and evaluation of logarithmic expressions generally involve using formulas directly or inversely. You should develop the habit of applying formulas in forward, backward, and deformed ways. lg 2+lg 5=1 is often used when calculating logarithmic values. At the same time, pay attention to The deformation of each part should be reduced to its simplest form.

Logarithmic operation PPT, part 4: thinking methods

How to solve logarithmic equations

Typical example Solve the following equation:

(1)1/2(lg x-lg 3)=lg 5-1/2lg(x-10);

(2)lg x+2log(10x)x=2;

(3)log_(x^2 "-" 1)(2x2-3x+1)=1.

Solution: (1) First, x in the equation should satisfy x>10; secondly, the original equation can be transformed into lg√(x/3)=lg5/√(x"-" 10),

∴√(x/3)=5/√(x"-" 10), that is, x2-10x-75=0.

Solve to get x=15 or x=-5 (discard),

After verification, x=15 is the solution of the original equation.

Logarithmic operation PPT, part 5: in-class drills

1.log248-log23=()

A.log244 B.2 C.4 D.-2

Analysis: Original formula=log248/3=log216=log224=4. Therefore, choose C.

Answer:C

2.log52·log425 is equal to ()

A.-1 B.1/2 C.1 D.2

Analysis: Original formula=lg2/lg5•lg25/lg4=lg2/lg5•2lg5/2lg2=1.

Answer:C

3.The value of (log_8 49)/(log_2 7) is ()

A.2 B.3/2 C.1 D.2/3

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