Question

Given that Log 10^2=0.301 and log10^3= 0.477, find the value of log10^6

Answer 1

Step-by-step explanation:

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

So let's just say that:

$x=log_ba$ and that $y=log_bc$

This means that: $b^x=a$ and that $b^y=c$.

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

$a*c=b^{x+y}$

This is due to the exponent identity that: $b^a*b^c=b^{a+c}$

So if you rewrite this in logarithmic form you get:

$log_b{ac}=x+y$

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

$log_b{ac} = log_ba+log_bc$

Answer 2

Answer: 0.778

This value is approximate

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Work Shown:

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

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