Logarithms help (1 Viewer)

[TearsOfFire]

Not sure if this is 2 unit or ext, but I’m stuck with these log questions. I can’t seem to find the log rule which is required to solve the questions. Any help/working out is appreciated. Thank in advance.

a) 3^2x - 9.3^x - 10 = 0
b) 4^x – 13(2^x) + 40 = 0
c) 5^2x – 28.5^x + 75 = 0

 

[Animals]

*Waits for Namu to pounce*

 

[lyounamu]

Not sure if this is 2 unit or ext, but I’m stuck with these log questions. I can’t seem to find the log rule which is required to solve the questions. Any help/working out is appreciated. Thank in advance.

a) 3^2x - 9.3^x - 10 = 0
b) 4^x – 13(2^x) + 40 = 0
c) 5^2x – 28.5^x + 75 = 0

a)

Let m = 3^x

so 3^2x - 9 . 3^x - 10 = m^2 - 9m -10 = (m-10)(m+1)
so m = 10 or m = -1
3^x = 10 or 3^x = -1
x = ln3(10) = ln (10)/ln(3)

b)

Let m = 2^x

so 4^x - 13 . 2^x + 40 = m^2 - 13m + 40 = (m-8)(m-5)
so m=8 or 5
i.e. 2^x = 8 or 2^x = 5

x = 3 or x = ln2(5) = ln5/ln2

c) Let m = 5^x

so 5^2x - 28 . 5^x + 75 = m^2 - 28m + 75 = (m-25)(m-3)
so 5^x = 25 or 3
x = 2 or x = ln5(3) = ln3/ln5

 

[Animals]

And there we have it.

 

[Chinmoku03]

Was the dot meant to be a multiplication sign...? I've been thinking it was a decimal all this time

 

[lyounamu]

Was the dot meant to be a multiplication sign...? I've been thinking it was a decimal all this time

Yep, sorry for the confusion. The dot is supposed to be a little above the "ground level".

 

[Graceofgod]

a)

Let m = 3^x

so 3^2x - 9 . 3^x - 10 = m^2 - 9m -10 = (m-10)(m+1)
so m = 10 or m = -1
3^x = 10 or 3^x = -1
x = ln3(10) = ln (10)/ln(3)

b)

Let m = 2^x

so 4^x - 13 . 2^x + 40 = m^2 - 13m + 40 = (m-8)(m-5)
so m=8 or 5
i.e. 2^x = 8 or 2^x = 5

x = 3 or x = ln2(5) = ln5/ln2

c) Let m = 5^x

so 5^2x - 28 . 5^x + 75 = m^2 - 28m + 75 = (m-25)(m-3)
so 5^x = 25 or 3
x = 2 or x = ln5(3) = ln3/ln5

Hmm, correct me if I am wrong. But minor error.

2^x = 5
log2(5) = x
etc. Not ln2(5)

 

[Chinmoku03]

Yep, sorry for the confusion. The dot is supposed to be a little above the "ground level".

No, not your answer. I thought the original question looked like it was asking someone to help solve a different question.

3^(2x) - (9.3)^x - 10 = 0 <- My interpretation

3^(2x) - 9*(3^x) - 10 = 0 <- What you solved

Hope my notation didn't lose everyone >.>;

 

[TearsOfFire]

Was the dot meant to be a multiplication sign...? I've been thinking it was a decimal all this time

The dot's a decimal, not multiplication sign, so substitution doesn't work? Or does it?

No, not your answer. I thought the original question looked like it was asking someone to help solve a different question.

3^(2x) - (9.3)^x - 10 = 0 <- My interpretation

3^(2x) - 9*(3^x) - 10 = 0 <- What you solved

Hope my notation didn't lose everyone >.>;

My bad, I didn't make it clear enough.

 

[Graceofgod]

Look at the question... It is pretty obviously meant to be a multiplication.

a) 3^2x - 9.3^x - 10 = 0
b) 4^x – 13(2^x) + 40 = 0
c) 5^2x – 28.5^x + 75 = 0

I think you should check the question again TearsofFire, you appear to be mistaken.

 

[lyounamu]

Hmm, correct me if I am wrong. But minor error.

2^x = 5
log2(5) = x
etc. Not ln2(5)

Yep, it should be written as log2(5). But I just couldn't be bothered to swap all that when I realised it. I am used to writing that - only when I type.

 

[TearsOfFire]

Look at the question... It is pretty obviously meant to be a multiplication.



I think you should check the question again TearsofFire, you appear to be mistaken.

I rechecked it just then. I've included the Q with proper formatting

The answers say:

a) x = log 3 (10)
b) x = 1.5 log 4 (5)
c) x = 2 or log 5 (3)

 

[Timothy.Siu]

yeah its obviously a multiplication...which is supported by those anwers

 

[bored of sc]

If it was a decimal you could still do a substitution though. If if was 9.3 instead of 9*3 then you could change it to 3*3.1. I don't know what I'm talking about.

 

[tommykins]

Look at the question... It is pretty obviously meant to be a multiplication.



I think you should check the question again TearsofFire, you appear to be mistaken.

The questions are fine.

 

[Graceofgod]

The questions are fine.

I realise this. Reread what I was talking about.

 

[tommykins]

回复: Re: Logarithms help

fair enough