Solved on Jan 20, 2024

Find the base $\mathrm{x}$ where $135_{10}=343_\mathrm{x}$

STEP 1

Assumptions
1. The subscript "10" in $135_{10}$ indicates that the number is in base 10 (decimal system).
2. The subscript "X" in $343_{X}$ indicates that the number is in base X, which we need to find.
3. The two numbers, $135_{10}$ and $343_{X}$ , represent the same value in their respective number systems.

STEP 2

First, we need to express the decimal number $135_{10}$ in terms of its base X equivalent, $343_{X}$ .
$135_{10} = 3 \cdot X^2 + 4 \cdot X^1 + 3 \cdot X^0$

STEP 3

Expand the equation using the properties of exponents, where $X^0 = 1$ .
$135 = 3X^2 + 4X + 3$

STEP 4

We now have a quadratic equation in terms of X. We need to solve for X.

STEP 5

To solve the quadratic equation, we can try to factor it if possible. Let's check if there are integer solutions by using the rational root theorem, which suggests that any rational solution, when X is an integer, is a factor of the constant term (3 in this case).

STEP 6

We list the factors of 3, which are ±1, ±3.

STEP 7

We test these factors to see if any of them satisfy the equation.

STEP 8

Test $X = 1$ :
$135 = 3(1)^2 + 4(1) + 3 = 3 + 4 + 3 = 10 \neq 135$

STEP 9

Test $X = -1$ :
$135 = 3(-1)^2 + 4(-1) + 3 = 3 - 4 + 3 = 2 \neq 135$

STEP 10

Test $X = 3$ :
$135 = 3(3)^2 + 4(3) + 3 = 3 \cdot 9 + 12 + 3 = 27 + 12 + 3 = 42 \neq 135$

STEP 11

Test $X = -3$ :
$135 = 3(-3)^2 + 4(-3) + 3 = 3 \cdot 9 - 12 + 3 = 27 - 12 + 3 = 18 \neq 135$

STEP 12

Since none of the integer factors of 3 satisfy the equation, we can conclude that the base X is not an integer. We need to solve the quadratic equation for X using other methods.

STEP 13

We can use the quadratic formula to solve for X:
$X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 3$ , $b = 4$ , and $c = 3$ .

STEP 14

Calculate the discriminant $b^2 - 4ac$ :
$\Delta = b^2 - 4ac = 4^2 - 4 \cdot 3 \cdot 3$

STEP 15

Calculate the value of the discriminant:
$\Delta = 16 - 36 = -20$

STEP 16

Since the discriminant is negative, there are no real solutions for X. This means that the base X cannot be a real number.

STEP 17

We must re-evaluate our earlier steps as the assumption that the base X is a real number has led to a contradiction. It's possible that we made a mistake in our calculations or assumptions.

STEP 18

Re-examine the original equation:
$135_{10} = 343_{X}$

STEP 19

Reconsider the possible values of X. Since X is a base, it must be an integer greater than 1.

STEP 20

Realize that there might have been a mistake in the testing of the factors. Let's re-test the factors with careful calculations.

STEP 21

Re-test $X = 3$ with careful calculation:
$135 = 3(3)^2 + 4(3) + 3 = 3 \cdot 9 + 4 \cdot 3 + 3 = 27 + 12 + 3 = 42$
This is incorrect, as noted before. It seems there was no mistake in the initial testing.

STEP 22

Re-test $X = 5$ :
$135 = 3(5)^2 + 4(5) + 3 = 3 \cdot 25 + 4 \cdot 5 + 3 = 75 + 20 + 3 = 98$
This is also not equal to 135.

STEP 23

Re-test $X = 7$ :
$135 = 3(7)^2 + 4(7) + 3 = 3 \cdot 49 + 4 \cdot 7 + 3 = 147 + 28 + 3 = 178$
This is greater than 135, which means we have overshot the value.

STEP 24

Re-test $X = 6$ :
$135 = 3(6)^2 + 4(6) + 3 = 3 \cdot 36 + 4 \cdot 6 + 3 = 108 + 24 + 3 = 135$
This is equal to 135.

STEP 25

We have found that $X = 6$ satisfies the equation $135 = 3X^2 + 4X + 3$ .

STEP 26

Therefore, the base X for which the statement $135_{10} = 343_{X}$ is true is 6.
The solution is $X = 6$ .

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Find the base x\mathrm{x}x where 13510=343x135_{10}=343_\mathrm{x}13510​=343x​

STEP 1

STEP 2

First, we need to express the decimal number 13510135_{10}13510​ in terms of its base X equivalent, 343X343_{X}343X​.13510=3⋅X2+4⋅X1+3⋅X0135_{10} = 3 \cdot X^2 + 4 \cdot X^1 + 3 \cdot X^013510​=3⋅X2+4⋅X1+3⋅X0

STEP 3

Expand the equation using the properties of exponents, where X0=1X^0 = 1X0=1.135=3X2+4X+3135 = 3X^2 + 4X + 3135=3X2+4X+3

STEP 4

We now have a quadratic equation in terms of X. We need to solve for X.

STEP 5

To solve the quadratic equation, we can try to factor it if possible. Let's check if there are integer solutions by using the rational root theorem, which suggests that any rational solution, when X is an integer, is a factor of the constant term (3 in this case).

STEP 6

We list the factors of 3, which are ±1, ±3.

STEP 7

We test these factors to see if any of them satisfy the equation.

STEP 8

Test X=1X = 1X=1:135=3(1)2+4(1)+3=3+4+3=10≠135135 = 3(1)^2 + 4(1) + 3 = 3 + 4 + 3 = 10 \neq 135135=3(1)2+4(1)+3=3+4+3=10=135

STEP 9

Test X=−1X = -1X=−1:135=3(−1)2+4(−1)+3=3−4+3=2≠135135 = 3(-1)^2 + 4(-1) + 3 = 3 - 4 + 3 = 2 \neq 135135=3(−1)2+4(−1)+3=3−4+3=2=135

STEP 10

Test X=3X = 3X=3:135=3(3)2+4(3)+3=3⋅9+12+3=27+12+3=42≠135135 = 3(3)^2 + 4(3) + 3 = 3 \cdot 9 + 12 + 3 = 27 + 12 + 3 = 42 \neq 135135=3(3)2+4(3)+3=3⋅9+12+3=27+12+3=42=135

STEP 11

Test X=−3X = -3X=−3:135=3(−3)2+4(−3)+3=3⋅9−12+3=27−12+3=18≠135135 = 3(-3)^2 + 4(-3) + 3 = 3 \cdot 9 - 12 + 3 = 27 - 12 + 3 = 18 \neq 135135=3(−3)2+4(−3)+3=3⋅9−12+3=27−12+3=18=135

STEP 12

Since none of the integer factors of 3 satisfy the equation, we can conclude that the base X is not an integer. We need to solve the quadratic equation for X using other methods.

STEP 13

We can use the quadratic formula to solve for X:X=−b±b2−4ac2aX = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}X=2a−b±b2−4ac​​where a=3a = 3a=3, b=4b = 4b=4, and c=3c = 3c=3.

STEP 14

Calculate the discriminant b2−4acb^2 - 4acb2−4ac:Δ=b2−4ac=42−4⋅3⋅3\Delta = b^2 - 4ac = 4^2 - 4 \cdot 3 \cdot 3Δ=b2−4ac=42−4⋅3⋅3

STEP 15

Calculate the value of the discriminant:Δ=16−36=−20\Delta = 16 - 36 = -20Δ=16−36=−20

STEP 16

Since the discriminant is negative, there are no real solutions for X. This means that the base X cannot be a real number.

STEP 17

We must re-evaluate our earlier steps as the assumption that the base X is a real number has led to a contradiction. It's possible that we made a mistake in our calculations or assumptions.

STEP 18

Re-examine the original equation:13510=343X135_{10} = 343_{X}13510​=343X​

STEP 19

Reconsider the possible values of X. Since X is a base, it must be an integer greater than 1.

STEP 20

Realize that there might have been a mistake in the testing of the factors. Let's re-test the factors with careful calculations.

STEP 21

Re-test X=3X = 3X=3 with careful calculation:135=3(3)2+4(3)+3=3⋅9+4⋅3+3=27+12+3=42135 = 3(3)^2 + 4(3) + 3 = 3 \cdot 9 + 4 \cdot 3 + 3 = 27 + 12 + 3 = 42135=3(3)2+4(3)+3=3⋅9+4⋅3+3=27+12+3=42This is incorrect, as noted before. It seems there was no mistake in the initial testing.

STEP 22

Re-test X=5X = 5X=5:135=3(5)2+4(5)+3=3⋅25+4⋅5+3=75+20+3=98135 = 3(5)^2 + 4(5) + 3 = 3 \cdot 25 + 4 \cdot 5 + 3 = 75 + 20 + 3 = 98135=3(5)2+4(5)+3=3⋅25+4⋅5+3=75+20+3=98This is also not equal to 135.

STEP 23

Re-test X=7X = 7X=7:135=3(7)2+4(7)+3=3⋅49+4⋅7+3=147+28+3=178135 = 3(7)^2 + 4(7) + 3 = 3 \cdot 49 + 4 \cdot 7 + 3 = 147 + 28 + 3 = 178135=3(7)2+4(7)+3=3⋅49+4⋅7+3=147+28+3=178This is greater than 135, which means we have overshot the value.

STEP 24

Re-test X=6X = 6X=6:135=3(6)2+4(6)+3=3⋅36+4⋅6+3=108+24+3=135135 = 3(6)^2 + 4(6) + 3 = 3 \cdot 36 + 4 \cdot 6 + 3 = 108 + 24 + 3 = 135135=3(6)2+4(6)+3=3⋅36+4⋅6+3=108+24+3=135This is equal to 135.

STEP 25

We have found that X=6X = 6X=6 satisfies the equation 135=3X2+4X+3135 = 3X^2 + 4X + 3135=3X2+4X+3.

STEP 26

Therefore, the base X for which the statement 13510=343X135_{10} = 343_{X}13510​=343X​ is true is 6.The solution is X=6X = 6X=6.

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Find the base $\mathrm{x}$ where $135_{10}=343_\mathrm{x}$

First, we need to express the decimal number $135_{10}$ in terms of its base X equivalent, $343_{X}$ .
$135_{10} = 3 \cdot X^2 + 4 \cdot X^1 + 3 \cdot X^0$

Expand the equation using the properties of exponents, where $X^0 = 1$ .
$135 = 3X^2 + 4X + 3$

Test $X = 1$ :
$135 = 3(1)^2 + 4(1) + 3 = 3 + 4 + 3 = 10 \neq 135$

Test $X = -1$ :
$135 = 3(-1)^2 + 4(-1) + 3 = 3 - 4 + 3 = 2 \neq 135$

Test $X = 3$ :
$135 = 3(3)^2 + 4(3) + 3 = 3 \cdot 9 + 12 + 3 = 27 + 12 + 3 = 42 \neq 135$

Test $X = -3$ :
$135 = 3(-3)^2 + 4(-3) + 3 = 3 \cdot 9 - 12 + 3 = 27 - 12 + 3 = 18 \neq 135$

We can use the quadratic formula to solve for X:
$X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 3$ , $b = 4$ , and $c = 3$ .

Calculate the discriminant $b^2 - 4ac$ :
$\Delta = b^2 - 4ac = 4^2 - 4 \cdot 3 \cdot 3$

Calculate the value of the discriminant:
$\Delta = 16 - 36 = -20$

Re-examine the original equation:
$135_{10} = 343_{X}$

Re-test $X = 3$ with careful calculation:
$135 = 3(3)^2 + 4(3) + 3 = 3 \cdot 9 + 4 \cdot 3 + 3 = 27 + 12 + 3 = 42$
This is incorrect, as noted before. It seems there was no mistake in the initial testing.

Re-test $X = 5$ :
$135 = 3(5)^2 + 4(5) + 3 = 3 \cdot 25 + 4 \cdot 5 + 3 = 75 + 20 + 3 = 98$
This is also not equal to 135.

Re-test $X = 7$ :
$135 = 3(7)^2 + 4(7) + 3 = 3 \cdot 49 + 4 \cdot 7 + 3 = 147 + 28 + 3 = 178$
This is greater than 135, which means we have overshot the value.

Re-test $X = 6$ :
$135 = 3(6)^2 + 4(6) + 3 = 3 \cdot 36 + 4 \cdot 6 + 3 = 108 + 24 + 3 = 135$
This is equal to 135.

We have found that $X = 6$ satisfies the equation $135 = 3X^2 + 4X + 3$ .

Therefore, the base X for which the statement $135_{10} = 343_{X}$ is true is 6.
The solution is $X = 6$ .