Question 38

If $$a^{x+y} = a^6$$ and x>y by 2 find x.

Solution

$$a^{x+y} = a^6$$ย 

or,$$\log_a\left(a^{\left(x+y\right)}\right)=\log_aa^6.$$(taking$$\log_a$$on both sides)

or,$$x+y=6$$.................................(1)

x>y by 2 means $$x-y=2$$..................(2)

add (1) and (2) :

2x=8 or, x=4.

Put x=4 in (1) :

y=2.

D is correct choice.


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