If log _{10} 2 = 0.3010 and log _{10} 7 = 0.8 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We are given $\log _{10} 2 = 0.3010$ and $\log _{10} 7 = 0.8451$.We need to find the value of $\log _{10} 2.8$.$\log _{10} 2.8 = \log _{10} \left(

Vedclass · Accepted Answer

$0.4471$

Vedclass · Answer

(A) We are given $\log _{10} 2 = 0.3010$ and $\log _{10} 7 = 0.8451$.We need to find the value of $\log _{10} 2.8$.$\log _{10} 2.8 = \log _{10} \left( \frac{28}{10} ight) = \log _{10} \left( \frac{2^2 	imes 7}{10} ight)$.Using the properties of logarithms,$\log(a 	imes b) = \log a + \log b$,$\log(a/b) = \log a - \log b$,and $\log(a^n) = n \log a$:$\log _{10} 2.8 = \log _{10} (2^2) + \log _{10} 7 - \log _{10} 10$.$= 2 \log _{10} 2 + \log _{10} 7 - 1$.$= 2(0.3010) + 0.8451 - 1$.$= 0.6020 + 0.8451 - 1$.$= 1.4471 - 1 = 0.4471$.

If $\log _{10} 2 = 0.3010$ and $\log _{10} 7 = 0.8451,$ then the value of $\log _{10} 2.8$ is

Similar Questions

If $\log _{x} y=100$ and $\log _{2} x=10,$ then the value of $y$ is

If $\log x = \frac{\log y}{2} = \frac{\log z}{5}$,then $x^{4} \cdot y^{3} \cdot z^{-2} = $

$\frac{\log 49 \sqrt{7} + \log 25 \sqrt{5} - \log 4 \sqrt{2}}{\log 17.5} = $

If $\frac{\log _{2} a}{2}=\frac{\log _{3} b}{3}=\frac{\log _{4} c}{4}$ and $a^{1 / 2} \cdot b^{1 / 3} \cdot c^{1 / 4}=24$,then:

Find $\log _{3 / 2} 3.375$

Vedclass Test Series

Exam Paper Generator

Online Exam Module