Log equation
- Thread starter Valentas
- Start date
Hello, Valentas!
\(\displaystyle \text{We have: }\;\log_2(4^x + 1) \;=\;\log_2(2^x) + \log_2(2^{x+1} - 6)\)
. . . . . . . . \(\displaystyle \log(4^x + 1) \;=\;\log_2\left[2^x(2^{x+1} - 6)\right]\)
. . . . . . . . \(\displaystyle \log(4^x + 1) \;=\;\log_2\left[2^{2x+1} - 6\cdot2^x\right]\)
\(\displaystyle \text{"Un-log": }\qquad\; 4^x + 1 \;=\;2^{2x+1} - 6\cdot2^x\)
. . . . . . . . . \(\displaystyle (2^2)^x + 1 \;=\;2\cdot2^{2x} - 6\cdot2^x\)
. . . . . . . . . . \(\displaystyle 2^{2x} + 1 \;=\;2\cdot2^{2x} - 6\cdot 2^x\)
\(\displaystyle \text{Simplify: }\;2^{2x} - 6\cdot2^x - 1 \;=\;0\)
\(\displaystyle \text{Let }\,u\,=\,2^x\!:\;\;u^2 - 6u - 1 \;=\;0\)
\(\displaystyle \text{Quadratic Formula: }\;u \;=\;\frac{6 \pm\sqrt{6^2 - 4(1)(-1)}}{2(1)} \;=\;\frac{6\pm\sqrt{40}}{2} \;=\;\frac{6\pm2\sqrt{10}}{2} \;=\;3 \pm\sqrt{10}\)
\(\displaystyle \text{Back-substitute: }\;2^x \:=\:3 + \sqrt{10}\;\;\text{ (since }2^x\text{ must be positive)}\)
\(\displaystyle \text{Take logs: }\;\ln(2^x) \;=\;\ln(3 + \sqrt{10}) \quad\Rightarrow\quad x\ln 2 \;=\;\ln(3 + \sqrt{10})\)
\(\displaystyle \text{Therefore: }\;x \;=\;\frac{\ln(3+\sqrt{10})}{\ln2}\)
\(\displaystyle \text{(What are the directions for the second problem?)}\)
\(\displaystyle \text{We have: }\;\log_2(4^x + 1) \;=\;\log_2(2^x) + \log_2(2^{x+1} - 6)\)
. . . . . . . . \(\displaystyle \log(4^x + 1) \;=\;\log_2\left[2^x(2^{x+1} - 6)\right]\)
. . . . . . . . \(\displaystyle \log(4^x + 1) \;=\;\log_2\left[2^{2x+1} - 6\cdot2^x\right]\)
\(\displaystyle \text{"Un-log": }\qquad\; 4^x + 1 \;=\;2^{2x+1} - 6\cdot2^x\)
. . . . . . . . . \(\displaystyle (2^2)^x + 1 \;=\;2\cdot2^{2x} - 6\cdot2^x\)
. . . . . . . . . . \(\displaystyle 2^{2x} + 1 \;=\;2\cdot2^{2x} - 6\cdot 2^x\)
\(\displaystyle \text{Simplify: }\;2^{2x} - 6\cdot2^x - 1 \;=\;0\)
\(\displaystyle \text{Let }\,u\,=\,2^x\!:\;\;u^2 - 6u - 1 \;=\;0\)
\(\displaystyle \text{Quadratic Formula: }\;u \;=\;\frac{6 \pm\sqrt{6^2 - 4(1)(-1)}}{2(1)} \;=\;\frac{6\pm\sqrt{40}}{2} \;=\;\frac{6\pm2\sqrt{10}}{2} \;=\;3 \pm\sqrt{10}\)
\(\displaystyle \text{Back-substitute: }\;2^x \:=\:3 + \sqrt{10}\;\;\text{ (since }2^x\text{ must be positive)}\)
\(\displaystyle \text{Take logs: }\;\ln(2^x) \;=\;\ln(3 + \sqrt{10}) \quad\Rightarrow\quad x\ln 2 \;=\;\ln(3 + \sqrt{10})\)
\(\displaystyle \text{Therefore: }\;x \;=\;\frac{\ln(3+\sqrt{10})}{\ln2}\)
\(\displaystyle \text{(What are the directions for the second problem?)}\)
soroban said:
This is the way the radicand must be. You are squaring the coefficient (-6),
and it must be shown as such, regardless that \(\displaystyle (6)^2 = (-6)^2.\)
Also, there are restrictions that must be checked in the given problem.
When a logarithm is taken of a number, this number you're taking
the log of must be positive.
The "answer" is a candidate, and it must be compared to any restrictions.
\(\displaystyle 4^x + 1\) is always positive, so it's okay.
\(\displaystyle 2^{x + 1} - 6\) will be negative for certain x-values, so the
candidate answer has to be taken into account with it.
It turns out that the answer that Soroban gave does not make\(\displaystyle 2^{x + 1} - 6\) nonpositive,
so that is in fact the (real) solution.
J
JeffM
Guest
I am presuming that lg has the base e. What in my youth was standard notation in the US for that function was ln. If some other base is intended, then the answer must adjusted appropriately. By the way, your English is quite good.Valentas said:
How to solve 2nd picture's f) and h) . I have no clue about the F) but getting the answers of h) but i get them positive. They must be negative. What's wrong?
[2lg(- x)]^(1/2) = lg[(x^2)^(1/2)] = lg(|x|).
lg(|x|) = [2lg(- x)]^(1/2) = {lg[(- x)^2]}^(1/2) = [lg(x^2)]^(1/2).
Square both sides.
[lg(|x|)]^2 = lg(x^2) = 2lg(|x|).
Divide both sides by lg|x|.
lg(|x|) = 2.
Delog.
So, |x| = e^2.
BUT x must be < 0 because lg(- x) was in one of the initial terms.
So, |x| = - x.
So x = - e^2.
Let's check the answer.
[2lg[- (-e^2)]^(1/2) = [2lg(e^2)]^(1/2) = {lg[(e^2)^2]}^(1/2) = [lg(e^4)]^(1/2) = 4^(1/2) = 2.
lg[{[-(e^2)]^2}^(1/2)] = lg[(e^4)^(1/2)] = lg(e^2) = 2.
It is also doable by considering it to be log base 10.
\(\displaystyle x \ = \ -10^2 \ = \ -100\)
Check:
\(\displaystyle \sqrt{2\log_{10}{[-(-100})]} \ = ? \ \log_{10}\sqrt{(-100)^2}\)
Eventually, after the steps are worked out, it becomes:
\(\displaystyle 2 \ = \ 2\)
\(\displaystyle x \ = \ -10^2 \ = \ -100\)
Check:
\(\displaystyle \sqrt{2\log_{10}{[-(-100})]} \ = ? \ \log_{10}\sqrt{(-100)^2}\)
Eventually, after the steps are worked out, it becomes:
\(\displaystyle 2 \ = \ 2\)
J
JeffM
Guest
Lookagain
Thank you for your comment. You are of course quite right that the exact answer is - (b[sup:2ws50pw7]2[/sup:2ws50pw7]), where b is the base of lg. So whether the answer is
- 100 or - e[sup:2ws50pw7]2[/sup:2ws50pw7] depends on whether the base is 10 or e.
My suggested answer is simply wrong if the base of lg is anything other than e. I had hoped that my initial warning would be sufficient to avoid misleading the student about that. It seemed to me that the student was confused about why the answer was negative and that my post needed to focus on making that point clear. But my answer may also have implied that the problem is not generically soluble, an implication that is of course false and may have misled the student in a different respect.
I am finding it very difficult to give answers that are maximally likely to be comprehensible to a confused student (particularly when it is probable that the student's native language is not English) and also mathematically exact. It would have been mathematically more exact to solve the problem using a generic base and thereby show that the answer is negative for any base whatsoever. The slight addition in abstractness, however, might have made the exposition less comprehensible.
Thank you for your comment. You are of course quite right that the exact answer is - (b[sup:2ws50pw7]2[/sup:2ws50pw7]), where b is the base of lg. So whether the answer is
- 100 or - e[sup:2ws50pw7]2[/sup:2ws50pw7] depends on whether the base is 10 or e.
My suggested answer is simply wrong if the base of lg is anything other than e. I had hoped that my initial warning would be sufficient to avoid misleading the student about that. It seemed to me that the student was confused about why the answer was negative and that my post needed to focus on making that point clear. But my answer may also have implied that the problem is not generically soluble, an implication that is of course false and may have misled the student in a different respect.
I am finding it very difficult to give answers that are maximally likely to be comprehensible to a confused student (particularly when it is probable that the student's native language is not English) and also mathematically exact. It would have been mathematically more exact to solve the problem using a generic base and thereby show that the answer is negative for any base whatsoever. The slight addition in abstractness, however, might have made the exposition less comprehensible.