Powers, Radicals, Logarithms practice questions — A-Level Maths

Use the product rule for logarithms: $${\log }_3(9x)={\log }_{3}9+{\log }_{3}x$$ Since $9=3^2$, we have ${\log }_{3}9=2$, and we are given ${\log }_{3}x=4$: $${\log }_3(9x)=2+4=6$$

$27=3^3$, so $3^{x+1}=3^3\Rightarrow x+1=3\Rightarrow x=2$.

Apply the quotient rule to merge the difference into a single logarithm: $${\log }_{2}80-{\log }_{2}5={\log }_2\frac{80}{5}$$ Simplify the argument and evaluate, since $2^4=16$: $${\log }_{2}16=4$$

${\displaystyle \frac{1}{\sqrt{3}-1}}\cdot {\displaystyle \frac{\sqrt{3}+1}{\sqrt{3}+1}}={\displaystyle \frac{\sqrt{3}+1}{3-1}}={\displaystyle \frac{\sqrt{3}+1}{2}}$.

${\log }_{4}8={\displaystyle \frac{{\log }_{2}8}{{\log }_{2}4}}={\displaystyle \frac{3}{2}}$.

$2^x<2^3\iff x<3$, i.e. $x\in (-\mathrm{\infty },3)$.

${\log }_{2}6+{\log }_2{\displaystyle \frac{8}{3}}={\log }_2\text{\hspace{0.17em}⁣}(6\cdot {\displaystyle \frac{8}{3}})={\log }_{2}16=4$.

Apply the power law to the first term. $$2{\log }_{5}10={\log }_5{10}^2={\log }_{5}100$$

Now use the subtraction (quotient) law. $${\log }_{5}100-{\log }_{5}4={\log }_5\frac{100}{4}={\log }_{5}25$$

Finally write $25$ as a power of $5$. $${\log }_{5}25={\log }_5{5}^2=2$$

${\log }_3(81)={\log }_3(3^4)=4$ and ${\log }_3(3)=1$. The difference is $4-1=3$.

$2^5=32$, so ${\log }_{2}32=5$.