Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 33004 by Rio Mike last updated on 09/Apr/18

$$howdoisolve$$$$forx$$$$a)2^{3-x}+2^x=6$$$$b){\left({log}_{3}x\right)}^2-6\left({log}_{3}x\right)+9=0$$

Commented by gunawan last updated on 09/Apr/18

$$\mathrm{a}.2^3.2^{-x}+2^x=6$$$$\mathrm{let}{2}^x=a$$$$8a^{-1}+a=6$$$$\frac{8}{a}+a=6$$$$a^2-6a+8=0$$$$(a-2)(a-4)=0$$$$a=2\to 2^x=2$$$$x=1$$$$a=4$$$$2^x=2^2\to x=2$$$$x=(1,2)$$

Commented by gunawan last updated on 09/Apr/18

$$\mathrm{b}.\mathrm{let}{\log }_{3}x=a$$$$a^2-6a+9=0$$$$(a-3)(a-3)=0$$$$a=3$$$${log}_{3}x=3$$$$x=27$$

Commented by abdo imad last updated on 09/Apr/18

$$b){\left({log}_{3}x\right)}^2-6\left({log}_{3}x\right)+9=0\iff {({log}_3\left(x\right)-3)}^2=0\iff$$$${log}_3\left(x\right)=3\iff \frac{lnx}{ln3}=3\iff ln\left(x\right)=ln\left(27\right)\iff x=27.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com