In a log jam...

[Guest]

I've got stuck solving some simultaneous equations...

log3(x) = y = log9(2x-1)

The first thing I did was to convert to base 10, giving

lg(x) / lg(3) = y, and
lg(2x-1) / lg(9) = y.

I then subtracted one from the other to give 0:

[2.lg(x) / lg(9)] - [(lg(2x-1) / lg(9)] = 0, which simplifies to give

lg(x^2 / 2x-1) / lg(9) = 0

Even assuming I haven't made any mistakes so far, this may be way off track because I have no idea where to go from here...

I'm also having problems with solving for the positive value of x satisfying the equation log2(x)=log4(x+6), which is a similar setup.

Can anyone help?[/i]

 

[soroban]

Hello, ben654!

$\displaystyle log_3(x)\:=\:log_9(2x\,-\,1)$

When I see two bases that are "related", I do a little gymnastics.

$\displaystyle \text{Let }\log_9(2x\,-\,1)\:=\:a\;\;\Rightarrow\;\;9^a\:=\:2x\,-\,1\;\;\Rightarrow\;\;(3^2)^a\:=\:2x\,-\,1$

$\displaystyle \;\;\;3^{2a}\:=\:2x\,-\,1\;\;\Rightarrow\;\;2a\:=\:\log_3(2x\,-\,1)\;\;\Rightarrow\;\;a\:=\:\frac{1}{2}\cdot\log_3(2x\,-\,1)$

The equation becomes: $\displaystyle \,\log_3(x)\;=\;\frac{1}{2}\cdot\log_3(2x\,-\,1)$

We have: $\displaystyle \,2\cdot\log_3(x)\;=\;\log_3(2x\,-\,1)\;\;\Rightarrow\;\;\log_3(x^2)\;=\;\log_3(2x\,-\,1)$

Then: $\displaystyle \,\log_3(x^2)\;=\;\log_3(2x\,-\,1)\;\;\Rightarrow\;\;x^2\:=\:2x\,-\,1\;\;\Rightarrow\;\;x^2\,-\,2x\,+\,1\:=\:0$

Therefore: $\displaystyle \,(x\,-\,1)^2\:=\:0\;\;\Rightarrow\;\;x\,=\,1$

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Changing to base 10 should have worked . . . but it's tricky.

\(\displaystyle \L\;\;\;\frac{\log(x)}{\log(3)}\;=\;\frac{\log(2x\,-\,1)}{\log(9)}\)

Clear denominators: $\displaystyle \,\log(9)\cdot\log(x)\;=\;\log(3)\cdot\log(2x\,-\,1)$

$\displaystyle \;\;$Note that: $\displaystyle \,\log(9)\:=\:\log(3^2)\:=\:2\cdot\log(3)$

The equation becomes: $\displaystyle \:2\cdot\log(3)\cdot\log(x)\;=\;\log(3)\cdot\log(2x\,-\,1)$

Divide by $\displaystyle \log(3):\;\;2\cdot\log(x)\;=\;\log(2x\,-\,1)$

We have: $\displaystyle \,\log(x^2)\:=\;\log(2x\,-\,1)\;\;\Rightarrow\;\;x^2\:=\:2x\,-\,1$ . . . and so on.