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Question Number 151162 by tabata last updated on 18/Aug/21

Commented by tabata last updated on 18/Aug/21

$$howcanitsolve$$

Commented by mr W last updated on 18/Aug/21

$$youseethesolutionx=1.$$$$ifyou{can}^{\prime }tsee,noonecanhelpyou.$$

Answered by Olaf_Thorendsen last updated on 18/Aug/21

$$6^x+1=8^x-{27}^{x-1}$$$$\bullet x=1\mathrm{and}x=2\mathrm{are}\mathrm{two}\mathrm{almost}$$$$\mathrm{obvious}\mathrm{solutions}.$$$$$$$$\mathrm{We}\mathrm{have}:$$$$2^x{3}^x+1={\left(2^x\right)}^3-\frac{{\left(3^x\right)}^3}{27}$$$$\mathrm{Let}a=2^x\mathrm{and}b=3^x.$$$$ab+1=a^3-\frac{b^3}{27}$$$$b^3+27ab+27-27a^3=0$$$$(b^3+pb+q=0)$$$$$$$$\mathrm{This}\mathrm{equation}\mathrm{may}\mathrm{be}\mathrm{viewed}\mathrm{as}\mathrm{a}$$$$\mathrm{cubic}\mathrm{in}b.$$$$$$$$\mathrm{Its}\mathrm{discriminant}\mathrm{is}:$$$${\mathrm{\Delta }}_b\left(a\right)=-4p^3-27q^2$$$${\mathrm{\Delta }}_b\left(a\right)=-4{\left(27a\right)}^3-27{(27-27a^3)}^2$$$${\mathrm{\Delta }}_b\left(a\right)=-{27}^3(4a^3+1-2a^3+a^6)$$$${\mathrm{\Delta }}_b\left(a\right)=-{27}^3{(1+a^3)}^2$$$$$$$$\mathrm{wich}\mathrm{is}\mathrm{negative}\mathrm{for}\mathrm{all}\mathrm{a}\mathrm{except}\mathrm{for}$$$$a=-1,\mathrm{when}\mathrm{it}\mathrm{is}\mathrm{zero}.\mathrm{This}\mathrm{does}\mathrm{not}$$$$\mathrm{correspond}\mathrm{to}\mathrm{a}\mathrm{real}\mathrm{value}\mathrm{of}x,\mathrm{hence}$$$$\mathrm{for}\mathrm{all}\mathrm{possible}\mathrm{values}\mathrm{of}a\mathrm{there}\mathrm{is}\mathrm{a}$$$$\mathrm{unique}\mathrm{real}b\mathrm{satisfying}\mathrm{the}\mathrm{polynomial}$$$$\mathrm{which}\mathrm{is}:$$$$$$$$b=3(a-1).$$$$$$$$\mathrm{By}\mathrm{definition}\mathrm{we}\mathrm{have}b=3^{{\log }_{2}a}.\mathrm{So}\mathrm{it}$$$$\mathrm{remains}\mathrm{to}\mathrm{solve}{3}^{{\log }_{2}a}=3(a-1).$$$$\mathrm{By}\mathrm{inspection}\mathrm{we}\mathrm{find}\mathrm{the}\mathrm{solutions}$$$$a=2\mathrm{and}a=4\mathrm{corresponding}\mathrm{to}x=1$$$$\mathrm{and}x=2.\mathrm{These}\mathrm{are}\mathrm{the}\mathrm{only}\mathrm{solutions},$$$$\mathrm{because}\mathrm{such}\mathrm{an}\mathrm{exponential}\mathrm{function}$$$$\mathrm{intersects}\mathrm{any}\mathrm{line}\mathrm{in}\mathrm{at}\mathrm{most}$$$$\mathrm{two}\mathrm{points}.$$

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