Question

Solve using log Help would be much appreciated

Answer 1

$10^{x-3.6}-7=41$

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

$10^{x-3.6}-7=41$

Add 7 to both sides of the equation.

$10^{x-3.6}=48$

Take the logarithm of both sides of the equation.

$log(10^{x-3.6})=log(48)$

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

$(x-3.6)log(10)=log(48)$

Add 3.6 to both sides of the equation.

$x=log(48)-(-3.6)$

Solve by first finding $log(48)$ then subtract $(-3.6)$ from the answer:

1. $log(48)=1.6812412373755872181499834821531$

Now just solve the equation:

$1.6812412373755872181499834821531 - (-3.6)=5.2812412373755872181499834821531$

Your answer is C. $5.2812412373755872181499834821531$ or $5.2812$

Answer 2

Answer:

5.2812

Step-by-step explanation:

$10^{x-3.6}-7=41$

Add 7 to both sides:

$\implies 10^{x-3.6}=48$

Taking logs of base 10:

$\implies \log_{10}10^{x-3.6}=\log_{10}48$

Apply log rule  $\log a^b=b \log a$:

$\implies (x-3.6)\log_{10}10=\log_{10}48$

Apply log rule $\log_aa=1$:

$\implies (x-3.6)\times 1 =\log_{10}48$

Simplify and solve:

$\implies x-3.6=\log_{10}48$

$\implies x=\log_{10}48+3.6$

$\implies x = 5.281241237...$