e^x=243 and e^y=32 then e^(3x+4x)/5=?
Answers
Answer:
432
Explanation:
Given--->
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eˣ = 243 and eʸ = 32
To find --->There is a mistake in question
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I think its like this
e³ˣ ⁺ ⁴ʸ/ ⁵ =?
Solution---->Now
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eˣ = 243
We factorize 243
eˣ = 3×3×3×3×3
eˣ =3⁵
Now
eʸ = 32
We factorize 32 now
=2×2×2×2×2
eʸ =2⁵
e³ˣ⁺⁴ʸ/⁵ = e³ˣ/⁵ + ⁴ʸ/⁵
We have a law of exponent aᵐ⁺ⁿ =aᵐ aⁿ
applying this law here
= e³ˣ/⁵ e⁴ʸ/⁵
We have another law of exponent
( aᵐ) ⁿ =aᵐⁿ
= (eˣ )³/⁵ ( eʸ )⁴/⁵
= ( 3⁵ )³/⁵ ( 2⁵ )⁴/⁵
= 3³ 2⁴
= ( 3 × 3 × 3 ) ( 2× 2 ×2 ×2)
= ( 27 ) (16)
= 432
Additional information---->
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1)aᵐ / aⁿ = aᵐ⁻ⁿ
2)a⁰ = 1
3)aᵐ bᵐ = (ab)ᵐ
Answer:
432
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