Chapter 7 - Exponential and Logarithmic Functions - 7-5 Exponential and Logarithmic Equations - Practice and Problem-Solving Exercises - Page 473: 34

$x=33$

Recall the definition of a logarithm (pg. 451): $\log_{b}{x}=y$ iff $b^y=x$ Applying this definition to our equation (with $b=10, y=2, x=3x+1$), we get: $10^2=3x+1$ $100=3x+1$ $3x=100-1$ $3x=99$ $x=33$ We confirm that the answer works: $\log_{10}(3\cdot 33+1)=2$ $\log_{10}(100)=2$ $\log_{10}(10^2)=2$ $2=2$