My attempt:$$\frac{\log(2x+1)-\log4}{1-\log(3x+2)}=1$$ $$\frac{\log(2x+1)-\log4}{\log10-\log(3x+2)}=\log10$$ $$\frac{\log\frac{(2x+1)}{4}}{\log\frac{10}{(3x+2)}}=\log10$$ $$\frac{\frac{(2x+1)}{4}}{\frac{10}{(3x+2)}}=10$$ $$\frac{(2x+1)}{4}=10\cdot\frac{10}{(3x+2)}$$(is this even correct ??) $$\frac{(2x+1)}{4}=\frac{100}{(3x+2)}$$ $$(2x+1)100=4(3x+2)$$ $$200x+100=12x+8$$ $$200x+100=12x+8$$ $$188x=-92$$ $$x=-23/47$$ The solution is 2.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ No, you can't do $\dfrac{\log a}{\log b}=\dfrac{a}{b}$ $\endgroup$– egregCommented Dec 7, 2013 at 18:57
-
$\begingroup$ Thought so... Then what can I do? $\endgroup$– user114141Commented Dec 7, 2013 at 18:59
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
$$\frac{\log(2x+1)-\log4}{1-\log(3x+2)}=1$$ becomes $$ \log(2x+1)-\log4=1-\log(3x+2) $$ that is $$ \log\frac{2x+1}{4}=\log\frac{10}{3x+2} $$ Can you go on from here? (I assume decimal logarithms.)
You can't go from
$$ \frac{\log a}{\log b}=\log c $$ to $$ \frac{a}{b}=c $$ Try some values to convince you about this.
$\begingroup$
$\endgroup$
2
$$\frac{\log(2x+1)-\log4}{1-\log(3x+2)}=1$$ $$\frac{\log(2x+1)/4}{\log 10/(3x+2)}=1$$ $$\log(2x+1)/4=\log 10/(3x+2)$$ $$(2x+1)/4=10/(3x+2)$$ $$(2x+1)(3x+2)=40$$ $$6x^2+7x-38=0$$ $$x_{1,2}=\frac{-7\pm31}{12}:x_1=2,x_2=-\frac{19}{6}$$ in real field only $x_1=2$ is a solution
-
$\begingroup$ Do you think that $-19/6$ is a solution? $\endgroup$– egregCommented Dec 7, 2013 at 19:17
-
$\begingroup$ No, it's not, because if you insert it in 2x+1 or in 3x+2, you get a negative number and logarithm of a negative nuber does not exsist, so a nuber must be >0 and ≠1. $\endgroup$ Commented Dec 7, 2013 at 19:54