Question

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

Answer 1

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.

Answer 2

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`