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Question Number 54875 by shaddie last updated on 13/Feb/19

$$\mathrm{Given}\mathrm{that}\frac{\log {(3\mathrm{x}+1)}^{2\mathrm{x}-1}}{\log (3\mathrm{x}+1)}=5,\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}.$$

Answered by kaivan.ahmadi last updated on 14/Feb/19

$$\log {(3\mathrm{x}+1)}^{2\mathrm{x}-1}=\log {(3\mathrm{x}+1)}^5\Rightarrow$$$${(3\mathrm{x}+1)}^{2\mathrm{x}-1}={(3\mathrm{x}+1)}^5\Rightarrow 2\mathrm{x}-1=5\Rightarrow$$$$\mathrm{x}=3$$

Answered by Otchere Abdullai last updated on 14/Feb/19

$$solution$$$$\frac{log{(3x+1)}^{2x-1}}{log(3x+1)}=5$$$$\Rightarrow \frac{(2x-1)log(3x+1)}{log(3x+1)}=5$$$$\Rightarrow 2x-1=5$$$$\Rightarrow 2x=5+1$$$$\Rightarrow 2x=6$$$$\Rightarrow \frac{2x}{2}=\frac{6}{2}$$$$\Rightarrow x=3$$

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