In a lottery game, a player picks 7 numbers from 1 to 40. How many different choices does the player have if order doesn't matte
1 answer:
You might be interested in
There are two ways you can go about this: I'll explain both ways.
<span>
</span><span>Solution 1: Using logarithmic properties
</span>The first way is to use logarithmic properties.
We can take the natural logarithm to all three terms to utilise our exponents.
Hence, ln2ᵃ = ln5ᵇ = ln10ⁿ becomes:
aln2 = bln5 = nln10.
What's so neat about ln10 is that it's ln(5·2).
Using our logarithmic rule (log(ab) = log(a) + log(b),
we can rewrite it as aln2 = bln5 = n(ln2 + ln5)
Since it's equal (given to us), we can let it all equal to another variable "c".
So, c = aln2 = bln5 = n(ln2 + ln5) and the reason why we do this, is so that we may find ln2 and ln5 respectively.
c = aln2; ln2 =
c = bln5; ln5 =
Hence, c = n(ln2 + ln5) = n( + )
Factorise c outside on the right hand side.
c = cn(
+ 
)
1 = n( + )
= +
= 
and thus, n =
<span>Solution 2: Using exponent rules
</span>In this solution, we'll be taking advantage of exponents.
So, let c = 2ᵃ = 5ᵇ = 10ⁿ
Since c = 2ᵃ, 2 =![\sqrt[a]{c}](https://tex.z-dn.net/?f=%20%5Csqrt%5Ba%5D%7Bc%7D%20)
= 
Then, 5 =
and 10 =
But, 10 = 5·2, so 10 =·
∴ = ·
= +
and n =
<span>
</span><span>Solution 1: Using logarithmic properties
</span>The first way is to use logarithmic properties.
We can take the natural logarithm to all three terms to utilise our exponents.
Hence, ln2ᵃ = ln5ᵇ = ln10ⁿ becomes:
aln2 = bln5 = nln10.
What's so neat about ln10 is that it's ln(5·2).
Using our logarithmic rule (log(ab) = log(a) + log(b),
we can rewrite it as aln2 = bln5 = n(ln2 + ln5)
Since it's equal (given to us), we can let it all equal to another variable "c".
So, c = aln2 = bln5 = n(ln2 + ln5) and the reason why we do this, is so that we may find ln2 and ln5 respectively.
c = aln2; ln2 =
c = bln5; ln5 =
Hence, c = n(ln2 + ln5) = n(
Factorise c outside on the right hand side.
c = cn(
1 = n(
and thus, n =
<span>Solution 2: Using exponent rules
</span>In this solution, we'll be taking advantage of exponents.
So, let c = 2ᵃ = 5ᵇ = 10ⁿ
Since c = 2ᵃ, 2 =
Then, 5 =
and 10 =
But, 10 = 5·2, so 10 =
∴
and n =