2 inequations but please solve em without logarithm,..cuz we didnt learned logarithm...but if must do with logarithm go ahead i understand a bit of logs
im sorry if i posted in bad section?really sorry..
2 inequations but please solve em without logarithm,..cuz we didnt learned logarithm...but if must do with logarithm go ahead i understand a bit of logs
im sorry if i posted in bad section?really sorry..
Usefor rendering exponentials in latexCode:base^{exponent}
1. It follows that if $\displaystyle \frac{1}{3^x+5} > \frac{1}{3^{x+1}-1}$ then $\displaystyle 3^x+5 < 3^{x+1}-1$. This is because when you divide by a fraction it's the same as multiplying by it's inverse. The smaller the denominator the larger the expression
By the laws of exponents $\displaystyle 3^{x+1} = 3 \cdot 3^x$
$\displaystyle 3^x+5 < 3 \cdot 3^x - 1$
Don't be scared by the exponent, you can treat it as normal
$\displaystyle 6 < 2 \cdot 3^x $
$\displaystyle 3 < 3^x$
In most cases you'd then need to use logarithms but this one doesn't need them because if the bases are the same then the exponents must be the same:
$\displaystyle 1 < x$
actually my proffesor said that's wrong...cuz i solved exactly like you first time...anyway thanks for trying...he said that we cant multiply with dominators cuz destroy inequality something like that anyway waiting for other responses
by the way second inequation i solved...you dont have to solve again
Hello, icefirez!
For the first one, your professor is right . . . up to a point.
$\displaystyle 1)\;\;\frac{1}{3^x+5} \;>\;\frac{1}{3^{x+1} - 1}$
I'd like to cross-multiply, but I can't do it freely if either denominator is negative.
We can see that $\displaystyle (3^x+5)$ is always positive.
But $\displaystyle (3^{x+1} - 1)$ could be negative.
When is it positive?
. . $\displaystyle 3^{x+1}-1 \:>\:0 \quad\Rightarrow\quad 3^{x+1} \:>\:1 \quad\Rightarrow\quad 3^{x+1}\:>\:3^0 \quad\Rightarrow\quad x+1 \:>\:0 \quad\Rightarrow\quad x \:>\:-1$
So, as long as $\displaystyle {\color{blue}x \,>\,-1}$. both denominators are positive. .[1]
Cross-multiply: .$\displaystyle 3^{x+1}-1 \;>\;3^x+5 \quad\Rightarrow\quad 3^{x-1} - 3^x \:>\:6$
Factor: .$\displaystyle 3^x(3-1) \:>\:6 \quad\Rightarrow\quad 3^x\cdot 2 \:>\:6 \quad\Rightarrow\quad 3^x \:>\:3 \quad\Rightarrow\quad x \:>\:1 $
This satisfies [1], so the soluton is: .$\displaystyle \boxed{x \:>\:1}$
And Soroban is correct ‘up to a point’!
Note that $\displaystyle 3^{x+1}-1<0$ for all $\displaystyle x<-1$.
Because $\displaystyle 3^x+5>0$ for all $\displaystyle x$ that means
$\displaystyle \frac{1}{3^x+5}>0>\frac{1}{3^{x+1}-1} $ for $\displaystyle x<-1$
So the solution $\displaystyle \boxed{( - \infty , - 1) \cup (1,\infty )}$