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Question Number 181217 by Mastermind last updated on 23/Nov/22

Answered by Rasheed.Sindhi last updated on 23/Nov/22

$$3^x{(2x+1)}^2=225=3^2{5}^2$$$$x=2\wedge 2x+1=5$$$$\Rightarrow x=2\wedge 2\left(2\right)+1=5$$$$\Rightarrow x=2\wedge 5=5$$$$\Rightarrow x=2$$

Commented by mr W last updated on 23/Nov/22

$$canwesolveiftheeqn.ise.g.$$$$3^x{(2x+1)}^2=200?$$

Commented by Rasheed.Sindhi last updated on 23/Nov/22

$$I{can}^{\prime }t\mathit{s}\mathit{i}\mathit{r},ifx\notin \mathbb{Z}$$

Commented by mr W last updated on 23/Nov/22

$$ithinkwecanapplylambertW.$$$$letmetrylater.$$

Commented by mr W last updated on 23/Nov/22

$$3^x{(2x+1)}^2=200$$$$3^x={\left(\frac{\sqrt{200}}{2x+1}\right)}^2$$$$3^{\frac{x}{2}}=\frac{\sqrt{200}}{4(\frac{x}{2}+\frac{1}{4})}$$$$3^{\frac{x}{2}+\frac{1}{4}}=\frac{\sqrt{200\sqrt{3}}}{4(\frac{x}{2}+\frac{1}{4})}$$$$(\frac{x}{2}+\frac{1}{4})\ln 3e^{(\frac{x}{2}+\frac{1}{4})\ln 3}=\frac{\left(\ln 3\right)\sqrt{200\sqrt{3}}}{4}$$$$\Rightarrow x=\frac{2}{\ln 3}W\left[\frac{\left(\ln 3\right)\sqrt{200\sqrt{3}}}{4}\right]-\frac{1}{2}$$$$\approx \frac{2\times 1.3393676}{\ln 3}-\frac{1}{2}=1.93829$$

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