Need to find x. .5(1+e)^(-4.787+1.55x)=e^(-4.787+1.55x)

$$\\0.5(1+e)^{-4.787+1.55x}=e^{-4.787+1.55x}\\\\
$let $t=-4.787+1.55x\\\\
0.5(1+e)^{t}=e^{t}\\\\
ln(0.5(1+e)^{t})=ln(e^{t})\\\\
ln(0.5)+ln((1+e)^{t})=ln(e^{t})\\\\
ln(0.5)+tln(1+e)=t\\\\
ln(0.5)=t-tln(1+e)\\\\
t=ln(0.5)\\\\
t=\frac{ln(0.5)}{(1-ln(1+e))}\\\\
-4.787+1.55x=\frac{ln(0.5)}{(1-ln(1+e))}\\\\
1.55x=\frac{ln(0.5)}{(1-ln(1+e))}+4.787\\\\$$

 

$$\\x=\left[\frac{ln(0.5)}{(1-ln(1+e))}+4.787\right]/1.55\\\\$$

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$${\mathtt{5}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{e}}\right)}^{\left({\mathtt{\,-\,}}{\mathtt{4.787}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.55}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)} = {{\mathtt{e}}}^{\left({\mathtt{\,-\,}}{\mathtt{4.787}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.55}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)} \Rightarrow {{\mathtt{e}}}^{\left({\frac{\left({\mathtt{1\,550}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{4\,787}}\right)}{{\mathtt{1\,000}}}}\right)} = {\mathtt{5}}{\mathtt{\,\times\,}}{\left({\mathtt{e}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}^{\left({\frac{\left({\mathtt{1\,550}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{4\,787}}\right)}{{\mathtt{1\,000}}}}\right)} \Rightarrow {{\mathtt{e}}}^{\left({\frac{\left({\mathtt{1\,550}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{4\,787}}\right)}{{\mathtt{1\,000}}}}\right)} = {\mathtt{5}}{\mathtt{\,\times\,}}{\left({\mathtt{e}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}^{\left({\frac{\left({\mathtt{1\,550}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{4\,787}}\right)}{{\mathtt{1\,000}}}}\right)}$$

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