Answer:
[tex]\textsf{a)} \quad a=\dfrac{6b+11}{4}[/tex]
[tex]\textsf{b)} \quad x=6+2\sqrt{2}[/tex]
Step-by-step explanation:
Part (a)
Given equation:
[tex]\dfrac{4^a}{2^{3b}}=32\sqrt{2}[/tex]
Rewrite 4 as 2² and 32 as 2⁵ and √2 as [tex]\sf 2^{\frac{1}{2}}[/tex] :
[tex]\implies \dfrac{(2^2)^a}{2^{3b}}=2^5 \cdot 2^{\frac{1}{2}}[/tex]
[tex]\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:[/tex]
[tex]\implies \dfrac{2^{2a}}{2^{3b}}=2^5 \cdot 2^{\frac{1}{2}}[/tex]
[tex]\textsf{Apply exponent rule} \quad a^b \cdot a^c=a^{b+c}:[/tex]
[tex]\implies \dfrac{2^{2a}}{2^{3b}}=2^{\frac{11}{2}}[/tex]
[tex]\textsf{Apply exponent rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:[/tex]
[tex]\implies 2^{2a-3b}=2^{\frac{11}{2}}[/tex]
[tex]\textsf{Apply exponent rule} \quad a^{f(x)}=a^{g(x)} \implies f(x)=g(x):[/tex]
[tex]\implies 2a-3b=\dfrac{11}{2}[/tex]
Add 3b to both sides:
[tex]\implies 2a=3b+\dfrac{11}{2}[/tex]
Divide both sides by 2:
[tex]\implies a=\dfrac{6b+11}{4}[/tex]
Part (b)
Given equation:
[tex]3x=x\sqrt{2}+14[/tex]
Subtract x√2 from both sides:
[tex]\implies 3x-x\sqrt{2}=14[/tex]
Factor out x:
[tex]\implies x(3-\sqrt{2})=14[/tex]
Divide both sides by (3 - √2):
[tex]\implies x=\dfrac{14}{3-\sqrt{2}}[/tex]
Multiply the numerator and the denominator by the conjugate of the denominator (3 - √2):
[tex]\implies x=\dfrac{14(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}[/tex]
[tex]\implies x=\dfrac{42+14\sqrt{2}}{9+3\sqrt{2}-3\sqrt{2}-2}[/tex]
[tex]\implies x=\dfrac{42+14\sqrt{2}}{7}[/tex]
[tex]\implies x=\dfrac{42}{7}+\dfrac{14\sqrt{2}}{7}}[/tex]
[tex]\implies x=6+2\sqrt{2}[/tex]
