Question

Solve this please for x

Answer 1

the solution I have used the formulas of logarithmic functions that are

converting logarithmic form into exponential form

if log of why with base x

is less than equals to a

then

y= x^a equation 1st

log x is well defined only when

x >0 it forms equation II

$log_{3}(x - 2) \leqslant 2 \\ \implies \: \: \:x - 2 \leqslant {3}^{2} \\ \implies \: \: x - 2 \leqslant 9 \\ \implies \: \: x \leqslant 11 \: \: \: - (1)$

$also \: x - 2 > 0 \\ \implies \: \: x > 2 \: \: \: - (2) \\ \\ \\ from \: (1)and(2) \\ we \: get \: \\ 2 < x \leqslant 11$

---> x€(2;11].....

Answer 2

Answer:

$\log _3\left(x-2\right)\le \:2\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:2<x\le \:11\:\\ \:\mathrm{Interval\:Notation:}&\:(2,\:11]\end{bmatrix}$

Step-by-step explanation:

$\log _3\left(x-2\right)\le \:2$

$\gray{\mathrm{Apply\:log\:rule}:\quad \:a=\log _b\left(b^a\right)}$

$\gray{2=\log _3\left(3^2\right)}$

$\log _3\left(x-2\right)\le \log _3\left(3^2\right)$

$\gray{\mathrm{Simplify}}$

$\log _3\left(x-2\right)\le \log _3\left(9\right)$

$\gray{\mathrm{If\:}a>1\mathrm{,\:then\:}\log _a\left(f\left(x\right)\right)\le \:\log _a\left(g\left(x\right)\right)\mathrm{\:is\:equivalent\:to\:}f\left(x\right)\le \:g\left(x\right),\:f\left(x\right)>0}$

$\gray{a=3,\:f\left(x\right)=x-2,\:g\left(x\right)=9}$

$x-2\le \:9\quad \mathrm{and}\quad \:x-2>0$

$\black{x-2\le \:9\quad \mathrm{and}\quad \:x-2>0\quad :}$

$x-2\le \:9\quad \mathrm{and}\quad \:x-2>0$

$\gray{x-2\le \:9\quad :\quad x\le \:11}$

$\gray{x-2>0\quad :\quad x>2}$

$\gray{\mathrm{Combine\:the\:intervals}}$

$x\le \:11\quad \mathrm{and}\quad \:x>2$

$\gray{\mathrm{Merge\:Overlapping\:Intervals}}$

$2<x\le \:11$