Question

Hello i need help on these two questions thank you

Answer 1

Answer:

Question 5. $a=2$

Question 6. $d=4$

Step-by-step explanation:

Question 5.

Given:

$\frac{3^{a}2\sqrt{9}}{27\sqrt{4}} = 1$

Solution:

$\frac{3^{a}2\sqrt{9}}{27\sqrt{4}} = 1$

$3^{a} 2\sqrt{9} =27\sqrt{4}$

$3^{a} =\frac{27\sqrt{4} }{2\sqrt{9} }$

$3^{a}=\frac{27\sqrt{2^{2} } }{2\sqrt{3^{2} } }$

$3^{a}=\frac{27\times 2}{2\times 3}$

$3^{a}=\frac{27}{3}$

$3^{a}=9$

$3^{a}=3^{2}$

Take log both side.

$log(3)^{a} =log(3)^{2}$

Simplify th above equation.

$a\times log(3)=2\times log(3)$

log(3) cancelled both side

$a=2$

Question 6.

Given:

$\frac{3^{d}\sqrt{5}}{3^{2}\sqrt{45}} = 3$

Solution:

$\frac{3^{d}\sqrt{5}}{3^{2}\sqrt{45}} = 3$

$3^{d}\sqrt{5} =3\times 9\sqrt{45}$

$3^{d} =\frac{27\sqrt{45} }{\sqrt{5} }$

$3^{d} =27\sqrt{\frac{45}{5}}$

$3^{d} =27\times \sqrt{9}$

$3^{d} =27\times \sqrt{3^{2}}$

$3^{d} =27\times 3$

$3^{d} =81$

$3^{d}=3^{4}$

Take log both side.

$log(3)^{d} =log(3)^{4}$

Simplify the above equation.

$d\times log(3)=4\times log(3)$

log(3) cancelled both side

$d=4$

Therefore $a=2$ for question 5 and $d=4$ for question 6.