Methods Question

[secretweapon]

2log_e(x+2) - log_e(x) = log_e3(x-1)
For the above equations, i got answers as -2 and 4, but the answer only said 4. How does the -2 not work as a solution?
Thanks

[GMT. -_-]

When you plug in x= -2, loge3(x-1) and 2loge2(x+2) will not be defined.

[RuiAce]

When you plug in x= -2, loge3(x-1) and 2loge2(x+2) will not be defined.

(Nor will \(\log_e x\) actually - none of them will be defined)

[secretweapon]

When you plug in x= -2, loge3(x-1) and 2loge2(x+2) will not be defined.

What do you mean?
Also, can someone please give me a hint as to how to do this question? (don't solve it for me though!)
2^x + 18 * 2^-x = 11
Thanks

[RuiAce]

What do you mean?

He means, that if you try plugging -2 into the original equation, you get \(2 \log_e 0 - \log_e (-2) = \log_e (-9) \). All three of these expressions are nonsense because you can only take log of a strictly positive value. You never take log of 0 or a negative value.

[GMT. -_-]

What do you mean?
Also, can someone please give me a hint as to how to do this question? (don't solve it for me though!)
2^x + 18 * 2^-x = 11
Thanks

Think of multiplying everything by 2^x, then it will somewhat resemble a quadratic by letting something = a variable.

[secretweapon]

for the equation y = 3sin(2x) -4, with a domain of 0≤x≤pi/2
how come the domain over which the gradient is positive is [0, pi/4) and not (0, pi/4)?

[Lear]

for the equation y = 3sin(2x) -4, with a domain of 0≤x≤pi/2
how come the domain over which the gradient is positive is [0, pi/4) and not (0, pi/4)?

Check the gradient at 0 and observe whether it is positive or negative.

[secretweapon]

Check the gradient at 0 and observe whether it is positive or negative.

the number 0 doesn't have any gradient?

[Lear]

the number 0 doesn't have any gradient?

Wait what... I meant differentiate the function and sub in x=0 to find the instantaneous gradient there.

[secretweapon]

Wait what... I meant differentiate the function and sub in x=0 to find the instantaneous gradient there.

when i diff the function and let x = 6, it's equal to 6, so since it's 6, and 6>0, then it is [ instead of (
?
also, in a transformation matrix, can dilations and reflections be written in any order?

[Lear]

when i diff the function and let x = 6, it's equal to 6, so since it's 6, and 6>0, then it is [ instead of (
?
also, in a transformation matrix, can dilations and reflections be written in any order?

Wait is this Methods 1/2?
Quick lesson - If you differentiate a function and sub in a point, the instantaneous gradient at that point is either positive, negative or 0. If it is positive, that means the function is increasing at that very point. Negative means decreasing and 0 means it is stationary. Your question asked for the domain over which the gradient was positive. Differentiating your equation and subbing x=0 gives us a gradient of 6, which is POSITIVE. Therefore 0 must be included in the domain over which you specify the gradient is positive.

[secretweapon]

Wait is this Methods 1/2?
Quick lesson - If you differentiate a function and sub in a point, the instantaneous gradient at that point is either positive, negative or 0. If it is positive, that means the function is increasing at that very point. Negative means decreasing and 0 means it is stationary. Your question asked for the domain over which the gradient was positive. Differentiating your equation and subbing x=0 gives us a gradient of 6, which is POSITIVE. Therefore 0 must be included in the domain over which you specify the gradient is positive.

it's methods 3/4
so if the gradient is >0, it's [
and if the gradient is <0, it's (
and if the gradient is 0, what would it be?

[Lear]

I believe you don't include where gradient is 0. so (

[secretweapon]

I believe you don't include where gradient is 0. so (

Is the approach that you have demonstrated above the same when you find the domain over which the gradient is negative?, as in whether it's ) or ]
?