Question

need help with a math problem

Answer 1

\[m _{1}-m _{2}=25\log_{10}(b _{2}/b _{1})\]

Answer 2

@whpalmer4 how am i supposed to do this problem?

Answer 3

remember that \(\log (b/a) = \log b - \log a\)

Answer 4

yes i did that so far but i ended up doing something wrong

Answer 5

think of logs as exponents and it should be clear:

\[\log_{10} 10^3 = 3\]\[\log_{10} 10^7 = 7\]\[\log_{10}\frac{10^7}{10^3} = \log_{10}10^7 - \log_{10}10^3 = 7 -3 = 4\]\[\log_{10}\frac{10^7}{10^3} = \log_{10}10^4 = 4\]

Answer 6

why don't you show me your work?

\[m_{1}-m_{2}=25\log_{10}(b_{2}/b_{1})\]

Answer 7

i know that i'm supposed to get decimal (m1-m2) =log10(b2/b1)
10 decimal(m1-m2)=b2/b1
b1*10 decimal(m1-m2)b2 according to the books example

Answer 8

let's make life easier, and we'll assume (as most texts do) that \(\log_{10} x = \log x\)

\[\frac{m_1-m_2}{25} = \log (\frac{b_2}{b_1})\]Agreed?

Answer 9

also, I'm not sure what you mean by "decimal" here

Answer 10

\[\frac{m_1 - m_2}{25} = \log b_2 - \log b_1\]\[\frac{m_1-m_2}{25} + \log b_1 = \log b_2\]Raise the base of the logarithm (10) to each side:
\[10^{\frac{m_1-m_2}{25} + \log b_1} = b_2\]