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Find, to 3 significant figures, the value of x for which 5^x = 7 - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 2

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Find, to 3 significant figures, the value of x for which 5^x = 7. Solve the equation 5^2 - 12(5^x) + 35 = 0.

Worked Solution & Example Answer:Find, to 3 significant figures, the value of x for which 5^x = 7 - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 2

Step 1

Find, to 3 significant figures, the value of x for which 5^x = 7.

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Answer

To solve for x in the equation 5x=75^x = 7, we can take the logarithm of both sides. Using the logarithm properties, we have:

x=log7log5x = \frac{\log 7}{\log 5}

Calculating this gives:

  1. Find log70.8451\log 7 \approx 0.8451 and log50.6990\log 5 \approx 0.6990.
  2. Plug in these values:

x0.84510.69901.21x \approx \frac{0.8451}{0.6990} \approx 1.21

Thus, to three significant figures, the value of x is 1.21.

Step 2

Solve the equation 5^2 - 12(5^x) + 35 = 0.

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Answer

To solve the equation 5212(5x)+35=05^2 - 12(5^x) + 35 = 0, we can substitute y=5xy = 5^x.

This transforms our equation into:

y212y+35=0y^2 - 12y + 35 = 0

Next, we can factor this quadratic equation:

(y5)(y7)=0(y - 5)(y - 7) = 0

Thus, we have two potential solutions for y:

  1. y5=0y=5y - 5 = 0 \Rightarrow y = 5
  2. y7=0y=7y - 7 = 0 \Rightarrow y = 7

Since y=5xy = 5^x, we substitute back:

  1. For y=5y = 5, we have 5x=55^x = 5 which gives x=1x = 1.
  2. For y=7y = 7, we have 5x=75^x = 7, which will return to our previous equation to give x1.21x \approx 1.21.

Therefore, the solutions for the equation are x=1x = 1 and x1.21x \approx 1.21.

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