# Transformations and Modulus Question Help

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Alex159

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#1

Hi, I need help with part d of this question but ive attached all of the questions below too.

I set one side equal to 0 as they intersect at y = 0 but im not sure what to do after that. I assume you need to remove the ln but im not sure how.

My working/thought process:

I'm thinking after plotting the graphs they intersect at (2, 0) when a is 0. There is only one point of intersection. Therefore, I set the modulus side to 0 as y = 0 and then multiplied by e to get 0 = -e^(1/2x - a). I then set 1/2x - a to 0, making x = 2a. after subbing in (2,0) i know that when a is 0, x is 2. Then i get y = 2a + 2 but im not sure if thats right or if my method to get there is correct.

Thanks in advance

1. The curve y = ln x is transformed to the curve y = (1/2x - a) by means of translation followed by a stretch. It is given that a is a positive constant.

a) Give full details of the translation and stretch involved

b)Sketch y = ln(1/2x - a)

c) Sketch y = |ln(1/2x - a)|

d) State, in terms of a, the set of values of x for which |ln(1/2x - a)| = -ln(1/2x - a)

I set one side equal to 0 as they intersect at y = 0 but im not sure what to do after that. I assume you need to remove the ln but im not sure how.

My working/thought process:

I'm thinking after plotting the graphs they intersect at (2, 0) when a is 0. There is only one point of intersection. Therefore, I set the modulus side to 0 as y = 0 and then multiplied by e to get 0 = -e^(1/2x - a). I then set 1/2x - a to 0, making x = 2a. after subbing in (2,0) i know that when a is 0, x is 2. Then i get y = 2a + 2 but im not sure if thats right or if my method to get there is correct.

Thanks in advance

1. The curve y = ln x is transformed to the curve y = (1/2x - a) by means of translation followed by a stretch. It is given that a is a positive constant.

a) Give full details of the translation and stretch involved

b)Sketch y = ln(1/2x - a)

c) Sketch y = |ln(1/2x - a)|

d) State, in terms of a, the set of values of x for which |ln(1/2x - a)| = -ln(1/2x - a)

Last edited by Alex159; 1 month ago

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Muttley79

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#2

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Hi, I need help with part 4 of this question

I set one side equal to 0 as they intersect at y = 0 but im not sure what to do after that. I assume you need to remove the ln but im not sure how.

Thanks in advance

1. The curve y = ln x is transformed to the curve y = (1/2x - a) by means of translation followed by a stretch. It is given that a is a positive constant.

a) Give full details of the translation and stretch involved

b)Sketch y = ln(1/2x - a)

c) Sketch y = |ln(1/2x - a)|

d) State, in terms of a, the set of values of x for which |ln(1/2x - a)| = -ln(1/2x - a)

**Alex159**)Hi, I need help with part 4 of this question

I set one side equal to 0 as they intersect at y = 0 but im not sure what to do after that. I assume you need to remove the ln but im not sure how.

Thanks in advance

1. The curve y = ln x is transformed to the curve y = (1/2x - a) by means of translation followed by a stretch. It is given that a is a positive constant.

a) Give full details of the translation and stretch involved

b)Sketch y = ln(1/2x - a)

c) Sketch y = |ln(1/2x - a)|

d) State, in terms of a, the set of values of x for which |ln(1/2x - a)| = -ln(1/2x - a)

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Alex159

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Please post your working

**Muttley79**)Please post your working

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Muttley79

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edited the post with info

**Alex159**)edited the post with info

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BlueBazooka

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Hi, I need help with part d of this question but ive attached all of the questions below too.

I set one side equal to 0 as they intersect at y = 0 but im not sure what to do after that. I assume you need to remove the ln but im not sure how.

My working/thought process:

I'm thinking after plotting the graphs they intersect at (2, 0) when a is 0. There is only one point of intersection. Therefore, I set the modulus side to 0 as y = 0 and then multiplied by e to get 0 = -e^(1/2x - a). I then set 1/2x - a to 0, making x = 2a. after subbing in (2,0) i know that when a is 0, x is 2. Then i get y = 2a + 2 but im not sure if thats right or if my method to get there is correct.

Thanks in advance

1. The curve y = ln x is transformed to the curve y = (1/2x - a) by means of translation followed by a stretch. It is given that a is a positive constant.

a) Give full details of the translation and stretch involved

b)Sketch y = ln(1/2x - a)

c) Sketch y = |ln(1/2x - a)|

d) State, in terms of a, the set of values of x for which |ln(1/2x - a)| = -ln(1/2x - a)

**Alex159**)Hi, I need help with part d of this question but ive attached all of the questions below too.

I set one side equal to 0 as they intersect at y = 0 but im not sure what to do after that. I assume you need to remove the ln but im not sure how.

My working/thought process:

I'm thinking after plotting the graphs they intersect at (2, 0) when a is 0. There is only one point of intersection. Therefore, I set the modulus side to 0 as y = 0 and then multiplied by e to get 0 = -e^(1/2x - a). I then set 1/2x - a to 0, making x = 2a. after subbing in (2,0) i know that when a is 0, x is 2. Then i get y = 2a + 2 but im not sure if thats right or if my method to get there is correct.

Thanks in advance

1. The curve y = ln x is transformed to the curve y = (1/2x - a) by means of translation followed by a stretch. It is given that a is a positive constant.

a) Give full details of the translation and stretch involved

b)Sketch y = ln(1/2x - a)

c) Sketch y = |ln(1/2x - a)|

d) State, in terms of a, the set of values of x for which |ln(1/2x - a)| = -ln(1/2x - a)

I imagine using your graph sketches as you have done is a valid alternative route to take with this question instead btw, but I personally prefer the algebraic approach.

Reply here if you need help solving the equation.

Last edited by BlueBazooka; 1 month ago

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