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By taking logarithms of both sides, solve the equation $$4^{3p-1} = 5^{210}$$ giving the value of $p$ to one decimal place. - Edexcel - A-Level Maths Pure - Question 4 - 2020 - Paper 1

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By taking logarithms of both sides, solve the equation $$4^{3p-1} = 5^{210}$$ giving the value of $p$ to one decimal place.

Worked Solution & Example Answer:By taking logarithms of both sides, solve the equation $$4^{3p-1} = 5^{210}$$ giving the value of $p$ to one decimal place. - Edexcel - A-Level Maths Pure - Question 4 - 2020 - Paper 1

Step 1

Using logarithms on both sides

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Answer

Taking the logarithm of both sides of the equation yields:

extlog(43p1)=extlog(5210). ext{log}(4^{3p-1}) = ext{log}(5^{210}).

Utilizing the power rule of logarithms, we can express this as:

(3p1)extlog(4)=210extlog(5).(3p-1) ext{log}(4) = 210 ext{log}(5).

Step 2

Solving for $p$

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Answer

Rearranging the equation for pp gives:

(3p1)=210extlog(5)extlog(4)(3p-1) = \frac{210 ext{log}(5)}{ ext{log}(4)}

Thus,

3p=210extlog(5)extlog(4)+1,3p = \frac{210 ext{log}(5)}{ ext{log}(4)} + 1,

and dividing by 3 gives:

p=13(210extlog(5)extlog(4)+1).p = \frac{1}{3} \left( \frac{210 ext{log}(5)}{ ext{log}(4)} + 1 \right).

Step 3

Calculate $p$ to one decimal place

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Answer

Using a calculator to find the values:

  • Calculate log(5)0.6990\text{log}(5) \approx 0.6990
  • Calculate log(4)0.6021\text{log}(4) \approx 0.6021

Substituting these values into the equation:

p=13(210×0.69900.6021+1)27.1.p = \frac{1}{3} \left( \frac{210 \times 0.6990}{0.6021} + 1 \right) \approx 27.1.

Therefore, to one decimal place, the value of pp is approximately:

p27.1.p \approx 27.1.

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