Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Log Question

\log_{10} 8 = r

\log_{10} 9 = s

What is

\log_{10} 5

in terms of

r

and

s

?
[NO CALCULATOR]

Literally have no idea where to start with this question

(edited 5 years ago)

well 2 + 3 = 5

8 is related to 2

9 is related to 3

:flute:

:flute:

Assuming the question is instead log[base 10](08) = r
10^r = 8
10^s = 9

Does that help at all?

OP

Original post by FryOfTheMann

Assuming the question is instead log[base 10](08) = r
10^r = 8
10^s = 9

Does that help at all?

Original post by the bear

well 2 + 3 = 5

8 is related to 2

9 is related to 3

:flute:

:flute:

10^x = 10^{\dfrac{s}{2}}+10^{\dfrac{r}{3}}

Original post by Retsek

x

To pick up from the first response in this thread,

r = \log_{10}8 = \log_{10}2^{3} = 3\log_{10}2 \Rightarrow \log_{10}2= \frac{r}{3}

s = \log_{10}9 = \log_{10}3^{2} = 2\log_{10}3 \Rightarrow \log_{10}3 = \frac{s}{2}

Use

\log_{10}2

and

\log_{10}3

to find

\log_{10}5

in terms of

r

and

s

Original post by ManLike007

To pick up from the first response in this thread,

r = \log_{10}8 = \log_{10}2^{3} = 3\log_{10}2 \Rightarrow \log_{10}2= \frac{r}{3}

s = \log_{10}9 = \log_{10}3^{2} = 2\log_{10}3 \Rightarrow \log_{10}3 = \frac{s}{2}

Use

\log_{10}2

and

\log_{10}3

to find

\log_{10}5

in terms of

r

and

s

Got me scratching my head a bit as the "obvious" solution
log(5) = log(2 + 3)
does not work / is a common mistake.

So how about
log(5) = log(10/2) = 1 - log(2)

The log(9) part is possibly there to mislead?

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