[tex]\bf 3(1.05)^x+1=6\implies 3(1.05)^x=5\implies (1.05)^x=\cfrac{5}{3}\\\\
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log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\qquad thus\\\\
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log\left[ (1.05)^x \right]=log\left( \cfrac{5}{3} \right)\implies xlog\left[ (1.05) \right]=log\left( \cfrac{5}{3} \right)
\\\\\\
x=\cfrac{log\left( \frac{5}{3} \right)}{log(1.05)}[/tex]