If four times a certain number, increased by 6 is equal to 94, what is the number?
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Answer:
Answers are in bold type
Step-by-step explanation:
f(x) =
The parabola opens up, so has a minimum at the vertex.
Let (h, k) be the vertex
h = -b/2a = - (-144)/2(1) = 57
k = 57^2 - 144(57) = 3249 - 6498 = -3249
Therefore, the vertex is (57, -3249)
The minimum value is -3249
The domain is the set of real numbers.
The range = {y | y ≥ -3249}
The function decreases when -∞ < x < 57 and increases when 57 > x > ∞
The x - intercepts: = 0
x(x - 114x) = 0
x = 0 or x = 114
x-intercepts are (0, 0) and (0, 114)
When x = 0, then we get the y-intercept. So, 0^2 - 114(0) = 0
y-intercept is (0, 0)
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 =