[tex]\qquad \qquad\huge \underline{\boxed{\sf Answer}}[/tex]
Here we go ~
[tex]\qquad \sf \dashrightarrow \: 6 {}^{ - 2x + 1} = 36[/tex]
[tex]\qquad \sf \dashrightarrow \: {6}^{ - 2x + 1} = {6}^{2} [/tex]
now, let's apply logarithm on both sides with base (6)
[tex]\qquad \sf \dashrightarrow \: log_{6}( 6 {}^{ - 2x + 1} ) = log_{6}( {6}^{2} ) [/tex]
According to properties of logarithm, the exponent on the number come out and it's written as a product of exponent times the logarithm.
that is :
[tex] \sf [ log_{a}(b {}^{n} ) = n \times log_{ a }(b) ][/tex]
[tex]\qquad \sf \dashrightarrow \: ( - 2x + 1) \times log_{6}(6) = 2 \times log_{6}(6) [/tex]
now, as we know : when the base and argument of log are same, then value of log is 1
[tex]\qquad \sf \dashrightarrow \: - 2x + 1 = 2[/tex]
[tex]\qquad \sf \dashrightarrow \: - 2x = 2 - 1[/tex]
[tex]\qquad \sf \dashrightarrow \: - 2x = 1[/tex]
[tex]\qquad \sf \dashrightarrow \: x = - \dfrac{1}{2} [/tex]
