Question 58

If $$3^{a}=27^{b}=81^{c}$$ and $$abc=144$$, then the value of $$12(\frac{1}{a}+\frac{1}{2b}+\frac{1}{5c})$$ is:

Solution

$$3^{a}=27^{b}=81^{c}$$

$$3^{a}=3^{3b}=3^{4c}$$

a = 3b = 4c

$$abc=144$$

$$4c \times \frac{4c}{3} \times c=144$$

$$c^3 = 27$$

c= 3

$$12(\frac{1}{a}+\frac{1}{2b}+\frac{1}{5c})$$

$$12(\frac{1}{4c}+\frac{1}{8c/3}+\frac{1}{5c})$$

= $$12\frac{33}{40c}$$

= $$12 \times \frac{33}{40 \times 3} = \frac{33}{10}$$

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