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A number is written as 25 to number base x. twice the number is written as 52 to base x. find the value of x.

Mathematics

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Step-by-step solution

16 steps

1. Find the value of xx.

Step 1: Convert the numbers from base xx to base 10.
The number written as 25x25_x in base xx is equivalent to 2×x1+5×x0=2x+52 \times x^1 + 5 \times x^0 = 2x + 5 in base 10.
Twice the number is written as 52x52_x in base xx, which is equivalent to 5×x1+2×x0=5x+25 \times x^1 + 2 \times x^0 = 5x + 2 in base 10.

Step 2: Set up the equation based on the problem statement.
"Twice the number is written as 52x52_x" means 2×(2x+5)=5x+22 \times (2x + 5) = 5x + 2.

Step 3: Solve the equation for xx.

4x+10=5x+24x + 10 = 5x + 2

Subtract 4x4x from both sides:

10=x+210 = x + 2

Subtract 2 from both sides:

x=8x = 8

The value of xx must be greater than the digits used (5 and 2), so x=8x=8 is a valid base.
The value of xx is 8\boxed{\text{}8\text{}}

2. Evaluate the expressions to four significant figures.

Given 5=2.2361\sqrt{5} = 2.2361 and 7=2.6458\sqrt{7} = 2.6458.

Part a· 3 steps

(i). 72−37\sqrt{72} - \frac{3}{\sqrt{7}}

  1. Simplify the terms.

    We simplify 72\sqrt{72} as 36×2=62\sqrt{36 \times 2} = 6\sqrt{2}.
    We rationalize the denominator for 37\frac{3}{\sqrt{7}}: 37=377\frac{3}{\sqrt{7}} = \frac{3\sqrt{7}}{7}.
    The expression becomes 62−3776\sqrt{2} - \frac{3\sqrt{7}}{7}.
    Assumption: Since 2\sqrt{2} is not provided, but the problem asks for evaluation without tables/calculator using given values, we assume 2≈1.4142\sqrt{2} \approx 1.4142 (a commonly known approximation to similar precision as given values) is expected.

  2. Substitute the given and assumed values.

    62−377≈6×1.4142−3×2.645876\sqrt{2} - \frac{3\sqrt{7}}{7} \approx 6 \times 1.4142 - \frac{3 \times 2.6458}{7} ≈8.4852−7.93747\approx 8.4852 - \frac{7.9374}{7} ≈8.4852−1.133914...\approx 8.4852 - 1.133914...

  3. Perform the subtraction and round to four significant figures.

    ≈7.351285...\approx 7.351285...

    Rounding to four significant figures, we get 7.351\boxed{\text{}7.351\text{}}

Part b· 13 steps

(ii). 45+25\frac{4}{\sqrt{5}} + 2\sqrt{5}

  1. Simplify the expression by rationalizing the denominator.

    45+25=455+25\frac{4}{\sqrt{5}} + 2\sqrt{5} = \frac{4\sqrt{5}}{5} + 2\sqrt{5}

  2. Combine the terms.

    =(45+2)5=(45+105)5=1455= \left(\frac{4}{5} + 2\right)\sqrt{5} = \left(\frac{4}{5} + \frac{10}{5}\right)\sqrt{5} = \frac{14}{5}\sqrt{5} =2.85= 2.8\sqrt{5}

  3. Substitute the given value for 5\sqrt{5} and round to four significant figures.

    =2.8×2.2361= 2.8 \times 2.2361 =6.26108= 6.26108

    Rounding to four significant figures, we get 6.261\boxed{\text{}6.261\text{}}

    3. Evaluate log⁡2(a5bc2)\log_2 \sqrt{\left(\frac{a^5 b}{c^2}\right)}

    Given log⁡2a=3\log_2 a = 3, log⁡2b=−2\log_2 b = -2, and log⁡2c=5\log_2 c = 5.

  4. Rewrite the expression using logarithm properties.

    log⁡2(a5bc2)=log⁡2(a5bc2)1/2\log_2 \sqrt{\left(\frac{a^5 b}{c^2}\right)} = \log_2 \left(\frac{a^5 b}{c^2}\right)^{1/2}

    Using the power rule log⁡kMp=plog⁡kM\log_k M^p = p \log_k M:

    =12log⁡2(a5bc2)= \frac{1}{2} \log_2 \left(\frac{a^5 b}{c^2}\right)

  5. Apply the quotient and product rules for logarithms.

    Using the quotient rule log⁡kMN=log⁡kM−log⁡kN\log_k \frac{M}{N} = \log_k M - \log_k N:

    =12(log⁡2(a5b)−log⁡2(c2))= \frac{1}{2} (\log_2 (a^5 b) - \log_2 (c^2))

    Using the product rule log⁡k(MN)=log⁡kM+log⁡kN\log_k (MN) = \log_k M + \log_k N:

    =12((log⁡2a5+log⁡2b)−log⁡2c2)= \frac{1}{2} ((\log_2 a^5 + \log_2 b) - \log_2 c^2)

  6. Apply the power rule again.

    =12(5log⁡2a+log⁡2b−2log⁡2c)= \frac{1}{2} (5\log_2 a + \log_2 b - 2\log_2 c)

  7. Substitute the given values.

    =12(5(3)+(−2)−2(5))= \frac{1}{2} (5(3) + (-2) - 2(5)) =12(15−2−10)= \frac{1}{2} (15 - 2 - 10) =12(3)= \frac{1}{2} (3) =1.5= 1.5

    The value is 1.5\boxed{\text{}1.5\text{}}

    4. Find the mean of the new set of numbers.

    Given that the mean of x1,x2,x3,x4x_1, x_2, x_3, x_4 is 7.

  8. Use the definition of the mean to find the sum of the original numbers.

    x1+x2+x3+x44=7\frac{x_1 + x_2 + x_3 + x_4}{4} = 7 x1+x2+x3+x4=7×4=28x_1 + x_2 + x_3 + x_4 = 7 \times 4 = 28

  9. Set up the expression for the mean of the new set of numbers.

    The new numbers are 5+x1,5+x2,5+x3,5+x45 + x_1, 5 + x_2, 5 + x_3, 5 + x_4.

    Mean=(5+x1)+(5+x2)+(5+x3)+(5+x4)4\text{Mean} = \frac{(5 + x_1) + (5 + x_2) + (5 + x_3) + (5 + x_4)}{4}

  10. Simplify and substitute the sum of the original numbers.

    Mean=4×5+(x1+x2+x3+x4)4\text{Mean} = \frac{4 \times 5 + (x_1 + x_2 + x_3 + x_4)}{4} Mean=20+284\text{Mean} = \frac{20 + 28}{4} Mean=484\text{Mean} = \frac{48}{4} Mean=12\text{Mean} = 12

    The mean of the new numbers is 12\boxed{\text{}12\text{}}

    5. Calculate the mean age of the class.

    The frequency table is:
    Age (years) (xx) | 12 | 13 | 14 | 15
    Frequency (ff) | 8 | 10 | 4 | 2

  11. Calculate the sum of frequencies (∑f\sum f).

    ∑f=8+10+4+2=24\sum f = 8 + 10 + 4 + 2 = 24

  12. Calculate the sum of (age ×\times frequency) (∑fx\sum fx).

    ∑fx=(12×8)+(13×10)+(14×4)+(15×2)\sum fx = (12 \times 8) + (13 \times 10) + (14 \times 4) + (15 \times 2) ∑fx=96+130+56+30\sum fx = 96 + 130 + 56 + 30 ∑fx=312\sum fx = 312

  13. Calculate the mean age using the formula Mean=∑fx∑f\text{Mean} = \frac{\sum fx}{\sum f}.

    Mean=31224\text{Mean} = \frac{312}{24} Mean=13\text{Mean} = 13

    The mean age of the class is 13 years\boxed{\text{}13 \text{ years}\text{}}

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