Answer :

[tex]\$2500+\$600=3100\$\\\\100\%+6\%=106\%=1.06\\\\\$2500\cdot1.06^n=\$3100\ \ \ /:\$2500\\\\1.06^n=1.24\\\\log_{1.06}1.06^n=log_{1.06}1.24\\\\n=log_{1.06}1.24\\\\n\approx4\ (years)[/tex]

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