Logarithms: 2 log_3 X+4-log_0.3 X*log_3 X-2 log_0.3 X=0

[afkanonymous]

I've been stuck with this problem for a while.

2log₃X+4-log_0.3X*log₃X-2log_0.3X=0

The answes are supposed to be 0.09 and 1/9, but i need the steps. I can't do it myself, getting stuck everytime.
Your help would be awesome.

 

[ksdhart2]

Well, you say you're "get[ting] stuck everytime," which means you've tried it multiple times in multiple different ways, so you should have loads of work to show. Please share it with us, even though you know it's wrong, so we can see exactly where you're getting bogged down. The more specific you can be, the more helpful we can be. Thank you.

As a hint: \(\displaystyle \displaystyle log_{0.3}\left(x\right)=log_3\left(x\right)\cdot \frac{ln\left(3\right)}{ln\left(0.3\right)}\)

 

[pka]

2log₃X+4-log_0.3X*log₃X-2log_0.3X=0

I suspect that it should be \(\displaystyle 2\log_3(X+4)-\log_{0.3}(X)\cdot\log_3(X)-2\log_{0.3}(X)=0\)
In any case, \(\displaystyle \log_{0.3}(X)=\dfrac{\log(X)}{\log(0.3)}=\dfrac{ \log (X)}{ \log (3)-\log(10)}\)

Now that makes quite a different question. Does it not?

 

[Otis]

2*log₃(X) + 4 - log_0.3(X)*log₃(X) - 2*log_0.3(X) = 0

The [answers] are supposed to be [9/100] and 1/9 ...

I've added some symbols to your typing above, to help pka. (His interpretation has no real solution.)

First, change 0.3 to rational form.

Then see whether you can get log(X)/[log(3) - 1] from log_3/10(X), by using the change-of-base formula.

If you're still stuck after that, please follow ksdhart2's request and post your work.

 

[HallsofIvy]

One other point- two of your logarithms are to base 0.3 and one is to base 3. Is that correct or should they all be to base 0.3?

 

[lookagain]

One other point- two of your logarithms are to base 0.3
and one is to base 3. Is that correct or should they all be to base 0.3?

Those bases are correct. x = 9/100 or x = 1/9 with those bases as is.

 

[hoosie]

Logarithm Question Response


Logarithm Question
Solve this equation for X:

Information provided by student in original post:
2log₃X + 4 - log_0.3X*log₃X - 2log_0.3X=0
Answers: X = 0.09 or X = 1/9
​

​

2logX/log3 + 4 - logX/log0.3_*logX/log3 - 2logX/log0.3= 0​

Since log0.3​

= log(3/10)
= log3 - log10
= log3 - 1​

it follows:
​

2logX/log3 + 4 - logX/(log3 - 1)_*logX/log3 - 2logX/(log3 - 1) = 0​

Let y = logX

(2/log3)y + 4 - {1/(log3 - 1)}y * (1/log3)y - {2/(log3 - 1)}y = 0​

(2/log3)y + 4 - {1/(log3 - 1)}y * (1/log3)y - {2/(log3 - 1)}y = 0
[(2/log3) - 2/(log3 - 1)]y - [{1/(log3 - 1)}(1/log3)]y^2 + 4 = 0

Simplify y coefficient:

(2/log3) -2/(log3 - 1)
= {2(log3 -1) - 2log3} / {log3*(log3 - 1)}
= -2/{log3*(log3 - 1)}

Simplify y^2 coefficient:

- {1/(log3- 1)}(1/log3)
= -1/{log3*(log3- 1)}

Substituting these answers into our equation:


-[2/{log3*(log3 - 1)}]y - [1/{log3*(log3- 1)}]y^2 + 4 = 0
[1/{log3*(log3- 1)}]y^2 + [2/{log3*(log3 - 1)}]y - 4 = 0

Multiply both sides by log3*(log3 - 1):
y^2 + 2y - 4log3*(log3 - 1) = 0

Solve for y using quadratic formula:
y^2 + 2y - 4log3*(log3 - 1) = 0
y^2 + 2y + 0.99790625205759 = 0
Let a = 1, b = 2, c = 0.99790625205759
y = (1/2){-2 +- (4 - 4*1*0.99790625205759)^(1/2)}
= (1/2){-2 +- (4 - 3.99162500823036)^(1/2)}
= (1/2){-2 +- (0.00837499176964)^(1/2)}
= (1/2)(-2 +- 0.09151498112134)

Therefore:
y = (1/2)(-2 + 0.09151498112134)
= 1/2 * -1.90848501887866
= -0.95424250943933


or


y = (1/2)(-2 - 0.09151498112134)
= 1/2 * -2.09151498112134
= -1.04575749056067


Since y = logX we have:

log(X) = -0.95424250943933
or
log(X) = -1.04575749056067

X = 10^(-0.95424250943933)
= 0.11111111111111
= 1/9

or
X = 10^(-1.04575749056067)
= 0.09

Check your answers for y by using your graphics calculator to graph:

Y = [1/{log3*(log3- 1)}]y^2 + [2/{log3*(log3 - 1)}]y - 4

y - intercepts from graph:​

y = -0.95424250943933 ​

= -0.95424 rounded to 5 decimal places​

or ​

y = -1.04575749056067​

= -1.04576 rounded to 5 decimal places​

​

​

 

[stapel]

Logarithm Question
Solve this equation for X:

2log₃X + 4 - log_0.3X*log₃X - 2log_0.3X=0

Answers: X = 0.09 or X = 1/9
​

​

2logX/log3 + 4 - logX/log0.3_*logX/log3 - 2logX/log0.3= 0​

Since log0.3​

= log(3/10)
= log3 - log10
= log3 - 1​

it follows:
​

2logX/log3 + 4 - logX/(log3 - 1)_*logX/log3 - 2logX/(log3 - 1) = 0​

Let y = logX
​

Why are you introducing new variables? Why are you starting by assuming the solution values? What are you then "proving"?

Are you keeping the student's needs in mind?

 

[hoosie]

Why are you introducing new variables? Why are you starting by assuming the solution values? What are you then "proving"?

Are you keeping the student's needs in mind?

I am not starting my answer with the solution values - these were provided by the student which you would know if you had checked their post.
It makes sense to use a substitution y = logX as this shows the student that the answer can be found by solving a simple quadratic and checked against its parabolic graph. By providing the student with a step by step solution I hope to improve their understanding of how to solve an equation involving logarithms.

 

[stapel]

I am not starting my answer with the solution values - these were provided by the student which you would know if you had checked their post.

Yes, the student had been provided with the solutions. The student was hoping to learn how to find those solutions, not how to check them.

 

[hoosie]

Yes, the student had been provided with the solutions. The student was hoping to learn how to find those solutions, not how to check them.

I was not showing the student how to check the solutions as you say. I was clearly showing the student how to find the answers using a step by step approach. It would certainly have benefited his understanding of how to solve an equation involving logarithms - both algebraically and graphically. It is important a student understands when to use a y = logX substitution to solve a logarithmic equation. It is not a bad thing to introduce a new variable in this case.