Problem: If 3=k⋅2r3=k \cdot 2^{r}3=k⋅2r and 15=k⋅4r15=k \cdot 4^{r}15=k⋅4r, then r=r=r=
Answer Choices:
A. −25-\log _{2} 5−log25
B. 52\log _{5} 2log52
C. 105\log _{10} 5log105
D. 25\log _{2} 5log25
E. 52\dfrac{5}{2}25
Solution:
Since 3=k⋅2r3=k \cdot 2^{r}3=k⋅2r and 15=k⋅4r15=k \cdot 4^{r}15=k⋅4r, we have
5=153=k⋅4rk⋅2r=22r2r=2r5=\dfrac{15}{3}=\dfrac{k \cdot 4^{r}}{k \cdot 2^{r}}=\dfrac{2^{2 r}}{2^{r}}=2^{r} 5=315=k⋅2rk⋅4r=2r22r=2r
Thus, by definition, r=25r=\log _{2} 5r=log25.