Question

solve the equation. log 2x+ log x = 11 x = (round to the nearest whole number as needed.)

Explanation:

Step1: Apply log - product rule

Using the rule $\log a+\log b=\log(ab)$, we can rewrite the left - hand side of the equation $\log 2x+\log x = 11$ as $\log(2x\cdot x)=\log(2x^{2}) = 11$.

Step2: Convert from logarithmic to exponential form

If $\log y = z$, then $y = 10^{z}$ (assuming base - 10 logarithms). So, from $\log(2x^{2})=11$, we get $2x^{2}=10^{11}$.

Step3: Solve for $x^{2}$

Divide both sides of the equation $2x^{2}=10^{11}$ by 2: $x^{2}=\frac{10^{11}}{2}=5\times10^{10}$.

Step4: Solve for $x$

Take the square root of both sides. Since $x>0$ (because we have $\log x$ and $\log 2x$ in the original equation and the argument of a logarithm must be positive), $x=\sqrt{5\times10^{10}}=\sqrt{5}\times10^{5}$.
$x\approx2.24\times10^{5}=224000$.

Answer:

$224000$