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Solve (a) $s^5 = 8$, giving your answer to 3 significant figures, (b) $\log_2 (x + 1) - \log_2 x = \log_2 7.$ - Edexcel - A-Level Maths Pure - Question 5 - 2005 - Paper 2

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Solve--(a)-$s^5-=-8$,-giving-your-answer-to-3-significant-figures,--(b)-$\log_2-(x-+-1)---\log_2-x-=-\log_2-7.$-Edexcel-A-Level Maths Pure-Question 5-2005-Paper 2.png

Solve (a) $s^5 = 8$, giving your answer to 3 significant figures, (b) $\log_2 (x + 1) - \log_2 x = \log_2 7.$

Worked Solution & Example Answer:Solve (a) $s^5 = 8$, giving your answer to 3 significant figures, (b) $\log_2 (x + 1) - \log_2 x = \log_2 7.$ - Edexcel - A-Level Maths Pure - Question 5 - 2005 - Paper 2

Step 1

Solve $s^5 = 8$

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Answer

To solve the equation s5=8s^5 = 8, we first rewrite 8 as a power of 2:

8=238 = 2^3

Thus, we can equate the exponents:

s5=23s^5 = 2^3

Taking the fifth root of both sides gives:

s=23/5s = 2^{3/5}

Calculating this gives:

s20.61.5157s \approx 2^{0.6} \approx 1.5157

Rounding to three significant figures, we get:

s1.52s \approx 1.52

Step 2

Solve $\log_2 (x + 1) - \log_2 x = \log_2 7$

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Answer

To solve the equation, we can use the properties of logarithms. Recall that:

logablogac=logabc\log_a b - \log_a c = \log_a \frac{b}{c}

Thus, we can rewrite the left side:

log2x+1x=log27\log_2 \frac{x + 1}{x} = \log_2 7

By exponentiating both sides, we eliminate the logarithm:

x+1x=7\frac{x + 1}{x} = 7

This simplifies to:

x+1=7xx + 1 = 7x

Rearranging gives:

1=7xx1=6xx=161 = 7x - x \\ 1 = 6x \\ x = \frac{1}{6}

Thus, the solution for xx is:

x0.167x \approx 0.167

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