how to find the value of x for which this statement is true. 4^2x+1=8

You can show this using the calc on the forum.  That way we can see straight away if it is correct :)

$${{\mathtt{4}}}^{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{8}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\frac{{ln}{\left({\mathtt{\,-\,}}{\sqrt{{\mathtt{7}}}}\right)}}{{ln}{\left({\mathtt{4}}\right)}}}\\
{\mathtt{x}} = {\frac{{ln}{\left({\mathtt{7}}\right)}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{4}}\right)}\right)}}\\
\end{array} \right\}$$

$${\frac{{ln}{\left({\mathtt{7}}\right)}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{4}}\right)}\right)}} = {\mathtt{0.701\: \!838\: \!730\: \!514\: \!401}}$$

4^2x+1=8

4^2x=7

2x=log₄(7)

x=$$\frac{\log_4(7)}{2}$$

Thanks EinsteinJr

The web2 calc will calculate log base 4 but most will not.

This is a more usual approach

4^2x=7

log4^2x=log7

2x*log4=log7

2x=log7/log4

x=log7/(2*log4)

the log can be any base so long as they are both the same base .

So x≈0.701838730514401 (this isn't the exact value, the number being irrationnal)