Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

Solve using log
Help would be much appreciated


Solve Using Log Help Would Be Much Appreciated class=

Sagot :

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]