question about logarithm

I

inverse

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\(\displaystyle 3^{1-x}-3^x=2 \)


I want to resolve this logarithm ecuation, I've applied the next method, I'd want to know where is my mistake


\(\displaystyle ln3^{1-x}-ln3^x=ln2 \)


\(\displaystyle ln(3^{1-x})/(ln3^x)=ln2 \)


\(\displaystyle ln 3^{1-2x}=ln2 \)

\(\displaystyle (1-2x)ln3=ln2\)


\(\displaystyle -2x=(ln2/ln3)-1\)


\(\displaystyle x=0,18\)

Thank you very much
 
Your very first step is very bad. There is no Distributive Property of the Logarithms Function.

\(\displaystyle 3^{1-x} - 3^{x} = 3\cdot 3^{-x} - 3^{x} = \frac{3}{3^{x}} - 3^{x} = \frac{3 - 3^{2x}}{3^{x}}\)

Start with that. See if you can find a "quadratic" equation to help you on your way.
 
okay so I can pass \(\displaystyle 3^x\) to multiply with 0?

¿\(\displaystyle 3-3^{2x}=0\)?
 
Hello, inverse!

\(\displaystyle \text{Solve for }x\!:\;\;3^{1-x} - 3^x \:=\:2\)
Click to expand...

Multiply by \(\displaystyle 3^x\!:\;\;3 - 3^{2x} \:=\:2\!\cdot\!3^x \quad\Rightarrow\quad 3^{2x} + 2\!\cdot\!3^x - 3 \:=\:0 \)

Factor: .\(\displaystyle (3^x + 3)(3^x - 1) \:=\:0\)

We have: .\(\displaystyle 3^x + 3 \:=\:0 \quad \Rightarrow \quad 3^x \:=\:\text{-}3\;\;\text{ no real solution}\)
. . . . . . . .\(\displaystyle 3^x-1 \:=\:0 \quad\Rightarrow\quad 3^x \:=\:1 \quad\Rightarrow\quad x \:=\:0\)
 
Last edited:
Another way to do it: let \(\displaystyle y= 3^x\). Then \(\displaystyle 3^{-x}= \frac{1}{y}\).

Now \(\displaystyle 3^{1- x}- 3^x= 3(3^{-x})- 3^x= 3(\frac{1}{y})- y= 2\). Now the rest is the same as Soroban did: muliplying through by y, \(\displaystyle 3- y^2= 2y\) so \(\displaystyle y^2+ 2y= 3\). Adding 1 to both sides, \(\displaystyle y^2+ 2y+ 1= (y+1)^2= 4\). Taking the square root of both sides \(\displaystyle y+ 1= \pm 2\). So y+ 1= 2 which gives y= 1 or y+1= -2 which give y= -3. So \(\displaystyle y= 3^x= 1\), x= 0. Of course, as Soroban said, \(\displaystyle y= 3^x= -3\) is impossible. A positive real number to any real power is non-negative. The only solution is x= 0.
 
HallsofIvy said:
Another way to do it: let \(\displaystyle y= 3^x\). Then \(\displaystyle 3^{-x}= \frac{1}{y}\).

Now \(\displaystyle 3^{1- x}- 3^x= 3(3^{-x})- 3^x= 3(\frac{1}{y})- y= 2\). Now the rest is the same as Soroban did: muliplying through by y, \(\displaystyle 3- y^2= 2y\) so \(\displaystyle y^2+ 2y= 3\). Adding 1 to both sides, \(\displaystyle y^2+ 2y+ 1= (y+1)^2= 4\). Taking the square root of both sides \(\displaystyle y+ 1= \pm 2\). So y+ 1= 2 which gives y= 1 or y+1= -2 which give y= -3. So \(\displaystyle y= 3^x= 1\), x= 0. Of course, as Soroban said, \(\displaystyle y= 3^x= -3\) is impossible. A positive real number to any real power is non-negative. The only solution is x= 0.
Click to expand...
This is the best method.
 
Because he likes me! Or, perhaps because he believes that actually assigning a letter, like y, to something like \(\displaystyle 3^x\) makes it less confusing. We can't all be as brilliant as Soroban!
 
thank you very much soroban and HallsofIvy for your contribution. Now I know to solve equations of this type, but I still do not understand why I'm not using either logarithmic property,

\(\displaystyle \log_b \left ( \frac{x}{y} \right ) = \log_b(x) - \log_b(y) \,\)


¿\(\displaystyle log \left ( \frac{3^{1-x}}{3^x} \right ) \neq log 3^{1-x}-log3^x\)?
 
inverse said:
thank you very much soroban and HallsofIvy for your contribution.
Click to expand...

That really hurts.
Now I know to solve equations of this type, but I still do not understand why I'm not using either logarithmic property,

\(\displaystyle \log_b \left ( \frac{x}{y} \right ) = \log_b(x) - \log_b(y) \,\)

\(\displaystyle log \left ( \frac{3^{1-x}}{3^x} \right ) \neq log 3^{1-x}-log3^x\)?
Click to expand...
You may use loragithm properties whenever they are applicable.

1 + 9 = 10

By your usage of the logarithm, which is not applciable, we have

log(1) + log(9) = log(10)

0 + log(9) = 1 -- Which is quite false.

Start again

1 + 9 = 10

Correctly applying logarithms (although it doesn't get us anywhere)

log(1+9) = log(10)

log(10) = log(10)

1 = 1 -- Which should seem quite reaonsable.
 
tkhunny said:


You may use loragithm properties whenever they are applicable.

1 + 9 = 10

By your usage of the logarithm, which is not applciable, we have

log(1) + log(9) = log(10)

0 + log(9) = 1 -- Which is quite false.

Start again

1 + 9 = 10

Correctly applying logarithms (although it doesn't get us anywhere)

log(1+9) = log(10)

log(10) = log(10)

1 = 1 -- Which should seem quite reaonsable.
Click to expand...

aah, now I understand, I had not realized this.
Thank you very much tkhunny for your explanation, sorry for not naming you, I did not realize :(
 
inverse said:
Thank you very much tkhunny for your explanation, sorry for not naming you, I did not realize :(
Click to expand...

Really, I was joking. You just can't see the pleasant one-sided grin on my face.

If you keep asking questions, you'll get it. Way to hang in there!
 
inverse said:
thank you very much soroban and HallsofIvy for your contribution. Now I know to solve equations of this type, but I still do not understand why I'm not using either logarithmic property,

\(\displaystyle \log_b \left ( \frac{x}{y} \right ) = \log_b(x) - \log_b(y) \,\)


¿\(\displaystyle log \left ( \frac{3^{1-x}}{3^x} \right ) \neq log 3^{1-x}-log3^x\)?
Click to expand...

It is!! - that is:


\(\displaystyle log \left ( \frac{3^{1-x}}{3^x} \right ) \ = \ log \left ( 3^{1-x} \right ) \ - \ log \left (3^x \right ) \)

However, that is not relevant for your problem. In your problem, what you should know is:

\(\displaystyle log \left ( {3^{1-x}} - {3^x} \right ) \ \neq \ log 3^{1-x} - log3^x\)
 
inverse said:
thank you very much soroban and HallsofIvy for your contribution. Now I know to solve equations of this type, but I still do not understand why I'm not using either logarithmic property,

\(\displaystyle \log_b \left ( \frac{x}{y} \right ) = \log_b(x) - \log_b(y) \,\)


¿\(\displaystyle log \left ( \frac{3^{1-x}}{3^x} \right ) \neq log 3^{1-x}-log3^x\)?
Click to expand...
No, the last is a correct equation. And IF the problem had been \(\displaystyle \frac{3^{1-x}}{3^x}= 2\), you could have taken the log of both sides. But it wasn't, it was \(\displaystyle 3^{1-x}- 3^{x}= 2\). You cannot just 'put' logarithms into that.
 
I tried to search for all Denis' posts containing the word "best", but the system apparently refuses to compile such a long list.
 
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