Respuesta :

Answer: 0.778

This value is approximate

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Work Shown:

[tex]\log_{10}(2) \approx 0.301\\\\\log_{10}(3) \approx 0.477\\\\\log_{10}(6) = \log_{10}(2*3)\\\\\log_{10}(6) = \log_{10}(2)+\log_{10}(3) \ \text{ see note below}\\\\\log_{10}(6) \approx 0.301 + 0.477\\\\\log_{10}(6) \approx 0.778\\\\[/tex]

Note: I used the rule that log(A*B) = log(A)+log(B) which works for any valid log base.

edisonlara1212

Step-by-step explanation:

Since the question has already been answered, I'd like to add something new, and explain why: [tex]log_b(a*c)=log_ba+log_bc[/tex]

So let's just say that:

[tex]x=log_ba[/tex] and that [tex]y=log_bc[/tex]

This means that: [tex]b^x=a[/tex] and that [tex]b^y=c[/tex].

So if we were to multiply the two, a and c. You get

[tex]a*c=b^{x+y}[/tex]

This is due to the exponent identity that: [tex]b^a*b^c=b^{a+c}[/tex]

So if you rewrite this in logarithmic form you get:

[tex]log_b{ac}=x+y[/tex]

and remember what x and y are equal to? that's right, it's the logarithms

so now you substitute the logarithms back in and get

[tex]log_b{ac} = log_ba+log_bc[/tex]

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