Kernel Mbox To Pst Crack Fixed 🔅

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Kernel Mbox To Pst Crack Fixed 🔅



 
 
 
 
 
 
 

Kernel Mbox To Pst Crack

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Solving $ (1+2)^x = 8 $ through Taylor expansion

I’m trying to solve the equation $$(1+2)^x = 8 \qquad\text{for $x$ real.}$$
However, I’m already getting a bit messy and the Wolfram Alpha output is also wrong. My problem (that I haven’t been able to resolve by myself) is that the solution of the equation is $\exp(5.475)$ but the Taylor expansion around $x = 0$ gives $\exp(5.475-x)$. Why? Shouldn’t it be $\exp(5.475)$?

A:

$\exp(5.475-x)=\exp(x)\exp(5.475)=\exp(5.475)\exp(x)$
Thus $x=0$.

A:

The question is, why not start with the Taylor series?
$$(1+2)^x = \sum_{k=0}^{\infty} {{\frac{x^k}{k!}}(2)^k} = \sum_{k=0}^{\infty} {{\frac{x^k}{k!}}(2)^{k+1}}$$
and if we substitute for $k+1$ for $k$, i.e.
$$(1+2)^x = \sum_{k=0}^{\infty} {{\frac{x^k}{k!}}(2)^{k+1}} = \sum_{k=0}^{\infty} {{\frac{x^{k+1}}{k!}}(
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