Stanford Entry Test Question(Grade 12 maths)

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[odd trail]

Identify the possible values of x first. Then square both sides.

[paper vortex]

And then?

[odd trail]

Well, you'll only have one square root left, so isolate it on one side and square again.

And then it's easy.

[paper vortex]

Let me solve

[fresh obsidian]

first get rid of the square root for 4 then make the common denominator 2

then it will be easier to simplify

[paper vortex]

Right

(sq.(2x+20))^2 = (12-2sq.(x-5))^2

Basically this

[fresh obsidian]

now expand both sides and rearrange

[odd trail]

Oh. Not quite what I had in mind, but this will also work.

[paper vortex]

OHHHHHHH

I GOT IT LMAO

[paper vortex]

Speak your mind aswell please

[odd trail]

Well, I was thinking more in terms of (√((x + 10)/2) + √(x - 5))^2 = 36, but that doesn't matter that much.

[paper vortex]

Why doesn’t this method work?

[odd trail]

Uh... Well, because ln(x + y) is not ln(x) + ln(y).

[paper vortex]

oh. RIGHT It's ln(xy)

No but then still ln(xy) = ln(x) + ln(y)
so ln(0.5(x+10)) = ln(0.5) + ln(x+10)

[fresh obsidian]

yea that part is correct

[paper vortex]

so where is the wrong

[fresh obsidian]

but here we have the power

[paper vortex]

so?

[fresh obsidian]

From this line to here its wrong

[paper vortex]

second last line

[fresh obsidian]

Because that's like taking ln(a+b) = ln6
And you went to expand lna +lnb = ln6

[fresh obsidian]

Yes

[fresh obsidian]

So you can't progress much and simplify by taking ln on either side

[paper vortex]

tHIS EXPANSION IS CORRECT?

[fresh obsidian]

its better to expand $$\sqrt\frac{2x+20}{4} = 6-(x-5)^\frac{1}{2}$$

[lucid portal]

Huang

[paper vortex]

Okay but i dont get why i cant use ln

[fresh obsidian]

for ln? yes

[fresh obsidian]

becoz

[paper vortex]

okay

[fresh obsidian]

.

you cant do much with ln(a+b)=ln6

its a dead end

[paper vortex]

Wait let me process

[fresh obsidian]

k

brb

[paper vortex]

ln(a+b) cant be expanded into ln(a) + ln(b) though

okay 😭

[odd trail]

Why would you use logarithms here?

You square, isolate a root, then square again. Then it'll be a polynomial equation.

[paper vortex]

Yes yes I can, but there's no harm in exploring other ways

[odd trail]

Well, I guess. But I can't really see any other ways here.

[paper vortex]

Then the question comes, "why not this way?" Whats wrong

[fresh obsidian]

yes but thats what you did in ur working

[odd trail]

Well, as discussed above, you attempted to use a property ln(a + b) = ln(a) + ln(b), which is wrong.

[paper vortex]

whereeee

this?

[odd trail]

Yes.

[paper vortex]

but this is being multiplied? Its [0.5 x (x+10)]^0.5
and ik that ln(ab) = ln(a) + ln(b)

[fresh obsidian]

Isnt this what you did

You expanded and like multiplied the ln

[paper vortex]

OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

[fresh obsidian]

.

[paper vortex]

YEAH YEAH I GET IT NOW

[fresh obsidian]

The expansion of ln for this term is correct

[paper vortex]

Okay so what if

[fresh obsidian]

If we ignore the question

[fresh obsidian]

yea?

[paper vortex]

Yeah its a dead end

Doesn't work

[fresh obsidian]

yep

theres no point honestly

its better to just separate the variables and square both sides

[paper vortex]

Yeah it makes sense

+close

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