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Solve using log
Help would be much appreciated

Solve Using Log Help Would Be Much Appreciated class=

Sagot :

semsee45

Answer:

5.2812

Step-by-step explanation:

[tex]10^{x-3.6}-7=41[/tex]

Add 7 to both sides:

[tex]\implies 10^{x-3.6}=48[/tex]

Taking logs of base 10:

[tex]\implies \log_{10}10^{x-3.6}=\log_{10}48[/tex]

Apply log rule  [tex]\log a^b=b \log a[/tex]:

[tex]\implies (x-3.6)\log_{10}10=\log_{10}48[/tex]

Apply log rule [tex]\log_aa=1[/tex]:

[tex]\implies (x-3.6)\times 1 =\log_{10}48[/tex]

Simplify and solve:

[tex]\implies x-3.6=\log_{10}48[/tex]

[tex]\implies x=\log_{10}48+3.6[/tex]

[tex]\implies x = 5.281241237...[/tex]

[tex]10^{x-3.6}-7=41[/tex]

Use the rules of exponents and logarithms to solve the equation.

[tex]10^{x-3.6}-7=41[/tex]

Add 7 to both sides of the equation.

[tex]10^{x-3.6}=48[/tex]

Take the logarithm of both sides of the equation.

[tex]log(10^{x-3.6})=log(48)[/tex]

The logarithm of a number raised to a power is the power times the logarithm of the number.

[tex](x-3.6)log(10)=log(48)[/tex]

Add 3.6 to both sides of the equation.

[tex]x=log(48)-(-3.6)[/tex]

Solve by first finding [tex]log(48)[/tex] then subtract [tex](-3.6)[/tex] from the answer:

1. [tex]log(48)=1.6812412373755872181499834821531[/tex]

Now just solve the equation:

[tex]1.6812412373755872181499834821531 - (-3.6)=5.2812412373755872181499834821531[/tex]

Your answer is C. [tex]5.2812412373755872181499834821531[/tex] or [tex]5.2812[/tex]

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