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Answer 1

Given:

• ∛(3x-2)=4

• ⁵√(5x+2)=2

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To find:

• Value of x in each case

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Solution 1:

$\sf \longmapsto 3\sqrt{3x-2}=4$

Whenever we remove cube root we have to add exponent 1/3.

$\sf \longmapsto (3x-2)^{\frac{1}{3}}=4$

Now cubing both sides

$\sf \longmapsto (3x-2)^{\frac{1}{3}\times3}=4^3$

$\sf \longmapsto (3x-2)^{\frac{1}{\not 3}\times \not 3}=64$

$\sf \longmapsto (3x-2)^1=64$

$\sf \longmapsto 3x-2=64$

Now transport -2 from LHS to RHS.

$\sf \longmapsto 3x=64+2$

$\sf \longmapsto 3x=66$

Now divide 3 both sides

$\sf \longmapsto 3x\div 3=66\div 3$

$\Large\underline{\boxed{\sf x=22}}\star$

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Solution 2:

$\sf \longmapsto ^5\sqrt{5x+2}=2$

Whenever we have to remove root of 5, we have to add 1/5 as it's exponent.

$\sf \longmapsto (5x+2)^{\frac{1}{5}}=2$

Now multiply 5 in exponents in both LHS and RHS.

$\sf \longmapsto (5x+2)^{\frac{1}{5}\times 5}=2^5$

$\sf \longmapsto (5x+2)^{\frac{1}{\not 5}\times \not 5}=32$

$\sf \longmapsto (5x+2)^1=32$

$\sf \longmapsto 5x+2=32$

Now transport 2 from LHS to RHS

$\sf \longmapsto 5x=32-2$

$\sf \longmapsto 5x=30$

$\sf \longmapsto x=\dfrac{30}{5}$

$\Large\underline{\boxed{\sf x=6}}\star$

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Answer:-

(1.)  X=22

(2.) X=6

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