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$\begin{aligned} \log_{x+1} e^5 & = \log_x e^2 & \small \color{#3D99F6}{\text{Taking }\ln (\cdot) \text{ on both sides,}} \\ \frac 5{\ln(x+1)} & = \frac 2{\ln x} \\ 5 \ln x & = 2 \ln (x+1) \\ \implies x^5 & = (x+1)^2 \\ x^5 - (x+1)^2 & = 0 \end{aligned}$

Using Newton Raphson method , we find that $x \approx \boxed{1.425}$ .

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Newton Raphson method

$\begin{aligned} x_{n+1} & = x_n - \frac {f'(x_n)}{f(x_n)} = x_n - \frac {5x_n^4 - 2(x_n+1)}{x_n^5 - (x_n+1)^2} \\ x_0 & = 1.5 \\ x_1 & = 1.433846154 \\ x_2 & = 1.425426675 \\ x_3 & = 1.425299606 \end{aligned}$

I used an Excel spreadsheet to do the calculations.

$log_{x+1} e^5 = log_{x} e^2$

$ln (e^5) / ln(x+1) = ln (e^2) / ln(x)$

$5*ln(e) * ln(x) = 2* ln(e) * ln(x+1)$ but ln(e) = 1

$ln x^5 = (x+1)^2$ bringing coefficients inside the logs as powers

$x^5 = (x+1)^2$ since if log A = log B then A = B

Graph each side of the equation and find they cross at x = 1.425