Solve the problem,142 base (n) =47 base ten
To solve this problem, we need to find the value of the base, denoted as n, in which 142 in base n is equal to 47 in base ten.
Let's break down the representation of 142 base (n). In base n, the digits are multiplied by the corresponding power of n. So, in this case, it can be written as:
1 * (n^2) + 4 * (n^1) + 2 * (n^0) = 47
Simplifying the equation:
n^2 + 4n + 2 = 47
Rearranging:
n^2 + 4n - 45 = 0
Now, we need to solve this quadratic equation for n. Factoring or using the quadratic formula, we get:
(n + 9)(n - 5) = 0
This gives us two possible solutions: n = -9 or n = 5. However, since the base cannot be negative (bases are always positive), we discard n = -9.
Therefore, the solution to the problem is n = 5.
To solve the problem, we need to find the value of 'n' in base 10 that satisfies the equation 142 base (n) = 47 base ten.
First, let's convert the numbers to base 10:
142 base (n) = (1 * n^2) + (4 * n^1) + (2 * n^0)
To find the value of n, we need to set up an equation:
(1 * n^2) + (4 * n) + (2) = 47
Simplifying the equation, we have:
n^2 + 4n + 2 = 47
Rearranging the equation to the standard form of a quadratic equation, we get:
n^2 + 4n - 45 = 0
Now we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
n = (-4 ± √(4^2 - 4 * 1 * (-45))) / (2 * 1)
Simplifying further:
n = (-4 ± √(16 + 180)) / 2
n = (-4 ± √196) / 2
Now, we can solve for two possible values of n:
n₁ = (-4 + √196) / 2
n₂ = (-4 - √196) / 2
Simplifying these results, we have:
n₁ = (-4 + 14) / 2
n₂ = (-4 - 14) / 2
n₁ = 10 / 2
n₂ = -18 / 2
n₁ = 5
n₂ = -9
Since bases cannot be negative, the value of 'n' in base 10 that satisfies the equation is 5.
To solve the problem, we need to find the value of base (n) that corresponds to the base 10 number 47.
In a base (n) number system, the digits range from 0 to (n-1). So first, we convert the base (n) number 142 into base 10.
To do this, we multiply each digit by the corresponding power of the base (n) and add them together.
142 base (n) = (1 * n^2) + (4 * n^1) + (2 * n^0)
Now, we have to set this equation equal to the base 10 number 47:
(1 * n^2) + (4 * n^1) + (2 * n^0) = 47
To make it easier, I will convert the exponents to decimal form:
n^2 + 4n^1 + 2n^0 = 47
Simplifying the equation, we have:
n^2 + 4n + 2 = 47
Rearranging the terms, we get:
n^2 + 4n - 45 = 0
Now we have a quadratic equation in terms of n. To solve it, we can factorize or use the quadratic formula.
Factoring the equation, we have:
(n + 9)(n - 5) = 0
Setting each factor equal to zero, we get two possible values for n:
n + 9 = 0 -> n = -9
n - 5 = 0 -> n = 5
However, since the base of a number cannot be negative, we discard the value n = -9. Therefore, the answer is:
n = 5
So, in base 5, the number 142 is equivalent to the base 10 number 47.