Logarithmic equation, please help!
- Thread starter shervlad
- Start date
\(\displaystyle x^{\{2log_3(x) - 1.5x\}} = 10^{(1/2)}\) Is that the problem?
What have you tried, or do you need a hint to get started?
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Yes, I've tried two ways:
1st:
\(\displaystyle {\log _{10}}{x^{2{{\lg }^3}x - 1,5x}} = {1 \over 2}\)
and secound:
\(\displaystyle \left( {2{{\lg }^3}x - 1,5x} \right) \cdot \lg x = {1 \over 2}\)
but I don't know how to go further... any idea?
shervlad, can you answer JeffM's question?
Can you clarify the following:
1. 1,5 is ambiguous. Do you mean 1.5?
2. By lg, do you mean log or ln?
3. By \(\displaystyle lg^3x\) do you mean \(\displaystyle log_3x\) or \(\displaystyle log(3x)\).
They can all lead to different answers.
Cheers,
Sai.
Can you clarify the following:
1. 1,5 is ambiguous. Do you mean 1.5?
2. By lg, do you mean log or ln?
3. By \(\displaystyle lg^3x\) do you mean \(\displaystyle log_3x\) or \(\displaystyle log(3x)\).
They can all lead to different answers.
Cheers,
Sai.
Do you live in the US? 1.5 means \(\displaystyle \dfrac{3}{2}\) there. Is that what you mean by 1,5.
We do not know what you mean by lg. In the US \(\displaystyle log_e(x) = ln(x).\) Is that what you mean by lg?
Further, are you supposed to give an exact symbolic answer or an approximate numeric answer?
Is lg log base 10 for you or log base 2?
If it is log base 10, the wolfram alpha says there is one real solution, if it is log base 2 here are three real solutions.
In either case, your problem amounts to solving
\(\displaystyle \displaystyle 4z^4 - 3za^{\displaystyle z} - \lg(10)=0\)
Where \(\displaystyle a\) is your log's base and \(\displaystyle z=\lg(x)\).
What class is this for, anyway?
If it is log base 10, the wolfram alpha says there is one real solution, if it is log base 2 here are three real solutions.
In either case, your problem amounts to solving
\(\displaystyle \displaystyle 4z^4 - 3za^{\displaystyle z} - \lg(10)=0\)
Where \(\displaystyle a\) is your log's base and \(\displaystyle z=\lg(x)\).
What class is this for, anyway?
I live in Moldova (a small country in the east of Europe) and here 1,5 means \(\displaystyle {{3 \over 2}}\), \(\displaystyle \lg x\) means \(\displaystyle {\log _{10}}x\), and \(\displaystyle {\lg ^3}x\) means \(\displaystyle {\left( {\lg x} \right)^3}\) so, we can rewrite the equation in that way:
\(\displaystyle {x^{2{{\left( {{{\log }_{10}}x} \right)}^3} - {3 \over 2}x}} = \sqrt {10} \)
and yes, I'm supposed to give an exact numeric answer, an more precisely - 0.1 and 10, but I don't know how to get to this solution....
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Great. I know where Moldova is, and your English is quite good.
I have not yet worked out an answer, but the first thing I would do is look for substitutions to make this thing easier to work with.
\(\displaystyle Let\ y = log_{10}(x) \implies x = 10^y.\)
\(\displaystyle So\ x^{\left(2\left\{log_{10}(x)\right\}^3 - \frac{3x}{2}\right)} = \sqrt{10} \implies x^{\frac{1}{2}(4y^3 - 3*10^y)} = \sqrt{10} \implies x^{(4y^3 - 3 * 10^y)}= 10.\)
Now take logs. See whether this gets you anywhere.
EDIT: I think it leads you straight to daon's equation, which he seems to have solved. I'll let him take it from here because that equation is ugly.
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This is not an algebraic equation, and is not "solvable" which is why I asked which class this was for.
The answers given by OP also do not check out, so something is certainly wrong with the problem. There is in fact only one solution
http://www.wolframalpha.com/input/?i=solve+x^(2(log_10(x))^3-1.5x)+=+sqrt(10)+for+x
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I am in 10th grade in a music high school and I love math and I study it for myself. This exercise is from a book with exercises for tenth grade and it is not even from the hard category... I don't know... maybe there is a mistake....
In addition to my last post, I will note that the problem could have been asking you to solve this one:
\(\displaystyle \displaystyle x^{2\lg^3(x) - 1.5\lg(x)}=\sqrt{10}\)
In that case we have the following resulting polynomial, where again \(\displaystyle z=\lg(x)\):
\(\displaystyle 4z^4-3z^2-1=0\)
This is a quadratic in \(\displaystyle z^2\) and is solvable using the quadratic formula (or simply factoring). You will obtain real roots \(\displaystyle z=1,-1\) which give \(\displaystyle x=10, 1/10\) respectively.
shervlad,
The solution changes based on the base of your log. Since you confirmed that the base is 10, your solution is x ~= 0.24599.
You are solving a non-linear equation. So, you will have to solve it numerically.
If you have been exposed to numerical algorithms before, you can use the most simplest but efficient Newton-Raphson's scheme to solve for it by supplying an initial guess and letting it converge to the solution. I have a sample 'matlab' code (without any bells and whistles) for you below. But you can 'see through' the algorithm and program it in your language of choice.
Solution Process:
Cheers,
Sai.
The solution changes based on the base of your log. Since you confirmed that the base is 10, your solution is x ~= 0.24599.
You are solving a non-linear equation. So, you will have to solve it numerically.
If you have been exposed to numerical algorithms before, you can use the most simplest but efficient Newton-Raphson's scheme to solve for it by supplying an initial guess and letting it converge to the solution. I have a sample 'matlab' code (without any bells and whistles) for you below. But you can 'see through' the algorithm and program it in your language of choice.
Code:
clc;clear;
z=1; % Initial guess value
for i=1:1000
a=10; % Base value
f=4*z^4-3*z*a^z-log10(10); % loga(10). Change 'a' accordingly.
fp=16*z^3-(3+3*z)*a^z; % Derivative of 'f'
z=z-f/fp
end
f
z
x=a^z % Display value of the root or solution. Answer x=0.24599Solution Process:
Code:
The first 10 iterations look as follows:
z=1
z = 0.38636
z = -0.019266
z = -0.35491
z = -0.65190
z = -0.61797
z = -0.61053
z = -0.60930
z = -0.60912
z = -0.60909
z = -0.60909
------------
f = 1.6410e-005, function converges to zero.
z = -0.60909
------------
Thus, x = 0.24599Cheers,
Sai.
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Oh you are a genius my friend!!
That makes perfect sense!
These are the exact solutions that are given in the book, and the graph confirms it!
So those who wrote the exercise instead of 1.5logx they wrote just 1.5x... thank you very much! You made my life much easier!
