Point t is on line segment su. Given st=6 and tu=5 determine the length of su
2 answers:
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The remainder theorem tells you that when you do synthetic division with the 6, the remainder will be the same value as when you evaluate the function with the value of 6.
The answer is 15
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 =