Answer
$x=100\sqrt[3]{2500}\approx 1357.2088$
Work Step by Step
Recall the basic properties of logarithms (pg. 462):
Product Property:
$\log_b{mn}=\log_b{m}+\log_b{n}$
Quotient Property:
$\log_b{\frac{m}{n}}=\log_b{m}-\log_b{n}$
Power Property:
$\log_b{m^n}=n\log_b{m}$
The definition of a logarithm (pg. 451):
$\log_{b}{x}=y$ iff $b^y=x$
First, we apply the power property:
$\log x^3-\log 6+\log 2.4=9$
Next, we apply the quotient property:
$\log{\frac{x^3}{6}}+\log 2.4=9$
Now, we apply the product property:
$\log{\left(\frac{2.4x^3}{6}\right)}=9$
$\log_{10}{\left(0.4x^3\right)}=9$
Finally, we apply the definition of a logarithm:
$10^9=0.4x^3$
$\dfrac{10^9}{0.4}=x^3$
$2,500,000,000=x^3$
$\sqrt[3]{2,500,000,000}=x$
$10^2\sqrt[3]{2500}=x$
$x=100\sqrt[3]{2500}\approx 1357.2088$
We confirm that the answer works:
$3\log{1357.2088}-\log 6+\log 2.4=9$
$9.3979-0.7782+0.3802=9$
$9=9$
