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Ekponencijalna jednačina
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Ekponencijalna jednačina
od vasto » Nedelja, 02. Jun 2013, 23:22
[dispmath]\left(\frac{2}{3}\right)^x\left(\frac{3}{2}\right)^{x+1}\left(\frac{3}{4}\right)^{x-1}=2\frac{2}{3}[/dispmath]
Re: hitno ekponencijalna
od Daniel » Ponedeljak, 03. Jun 2013, 00:29
Nadam se da nije prekasno...
[dispmath]\left(\frac{2}{3}\right)^x\left(\frac{3}{2}\right)^{x+1}\left(\frac{3}{4}\right)^{x-1}=2\frac{2}{3}[/dispmath][dispmath]\left(\frac{3}{2}\right)^{-x}\left(\frac{3}{2}\right)^{x+1}\left(\frac{3}{4}\right)^{x-1}=\frac{6}{3}+\frac{2}{3}[/dispmath][dispmath]\left(\frac{3}{2}\right)^{-x+x+1}\left(\frac{3}{4}\right)^{x-1}=\frac{8}{3}[/dispmath][dispmath]\frac{3}{2}\cdot\left(\frac{3}{4}\right)^{x-1}=\frac{8}{3}[/dispmath][dispmath]\left(\frac{3}{4}\right)^{x-1}=\frac{8}{3}\cdot\frac{2}{3}[/dispmath][dispmath]\left(\frac{3}{4}\right)^{x-1}=\frac{16}{9}[/dispmath][dispmath]\left(\frac{4}{3}\right)^{1-x}=\left(\frac{4}{3}\right)^2[/dispmath][dispmath]1-x=2[/dispmath][dispmath]x=-1[/dispmath]
[dispmath]\left(\frac{2}{3}\right)^x\left(\frac{3}{2}\right)^{x+1}\left(\frac{3}{4}\right)^{x-1}=2\frac{2}{3}[/dispmath][dispmath]\left(\frac{3}{2}\right)^{-x}\left(\frac{3}{2}\right)^{x+1}\left(\frac{3}{4}\right)^{x-1}=\frac{6}{3}+\frac{2}{3}[/dispmath][dispmath]\left(\frac{3}{2}\right)^{-x+x+1}\left(\frac{3}{4}\right)^{x-1}=\frac{8}{3}[/dispmath][dispmath]\frac{3}{2}\cdot\left(\frac{3}{4}\right)^{x-1}=\frac{8}{3}[/dispmath][dispmath]\left(\frac{3}{4}\right)^{x-1}=\frac{8}{3}\cdot\frac{2}{3}[/dispmath][dispmath]\left(\frac{3}{4}\right)^{x-1}=\frac{16}{9}[/dispmath][dispmath]\left(\frac{4}{3}\right)^{1-x}=\left(\frac{4}{3}\right)^2[/dispmath][dispmath]1-x=2[/dispmath][dispmath]x=-1[/dispmath]
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