Question

solve each equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places:

${\log }_{16}x+{\log }_{4}x+{\log }_{2}x=7$

Short answer

The value of x satisfying the equation is$x=16$

Step-by-step solution

Given information

We are given an equation${\log }_{16}x+{\log }_{4}x+{\log }_{2}x=7$

Change the base 

We get,

$$\begin{gathered} {\log }_{16}x+{\log }_{4}x+{\log }_{2}x=7 \\ \frac{{\log }_2\left(x\right)}{{\log }_2\left(16\right)}+\frac{{\log }_2\left(x\right)}{{\log }_2\left(4\right)}+{\log }_2\left(x\right)=7 \\ \frac{{\log }_2\left(x\right)}{{\log }_2{\left(2\right)}^4}+\frac{{\log }_2\left(x\right)}{{\log }_2\left(2^2\right)}+{\log }_2\left(x\right)=7 \\ \frac{{\log }_2\left(x\right)}{4}+\frac{{\log }_2\left(x\right)}{2}+{\log }_2\left(x\right)=7 \\ {\log }_2\left(x\right)+2{\log }_2\left(x\right)+4{\log }_2\left(x\right)=28 \\ 7{\log }_2\left(x\right)=28 \\ {\log }_2\left(x\right)=4 \end{gathered}$$

Solve the equation

We get,

$$\begin{gathered} {\log }_2\left(x\right)=4 \\ Use:y={\log }_a\left(x\right)\iff x=a^y \\ x=2^4 \\ x=16 \end{gathered}$$

Conclusion

The value of x satisfying the equation$x=16$.