Log2(8x-3)-log2 5 =1

To solve the equation ${\log }_2(8x-3)-{\log }_{2}5=1$, we will use the properties of logarithms and follow a step-by-step algebraic approach. 1. Identify Constraints (Domain) The argument of a logarithm must be strictly greater than zero. Therefore: $8x-3>0$ $8x>3$ $x>\frac{3}{8}\text{(or}x>0.375)$ 2. Apply Logarithmic Properties Using the quotient rule, ${\log }_b\left(a\right)-{\log }_b\left(c\right)={\log }_b\left(\frac{a}{c}\right)$ , we can combine the left side of the equation: ${\log }_2\left(\frac{8x-3}{5}\right)=1$ 3. Convert to Exponential Form By definition, if ${\log }_b\left(y\right)=x$, then $b^x=y$. Applying this here: $\frac{8x-3}{5}=2^1$ $\frac{8x-3}{5}=2$ 4. Solve for $x$ Now, we perform basic algebraic operations to isolate $x$:

• Multiply both sides by 5:
  $8x-3=2\cdot 5$ $8x-3=10$ Add 3 to both sides:
  $8x=13$ Divide by 8:
  $x=\frac{13}{8}$ $x=1.625$

5. Verify the Result Since $1.625>0.375$, the solution falls within the defined domain. We can also plug it back into the original equation:

1. $8\left(1.625\right)-3=13-3=10$ ${\log }_2\left(10\right)-{\log }_2\left(5\right)={\log }_2(10/5)={\log }_2\left(2\right)$ ${\log }_2\left(2\right)=1$

Final Answer: $x=\frac{13}{8}\text{or}1.625$ Would you like me to solve another logarithmic equation or demonstrate how to apply these properties to a different base?