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  1. Mistake had occurred in step 1): it should be LHS =4^[5(x-3)]=2^[2*5(x-3)]=2^[10(x-3)] RHS=32^(2x)=2^[5*(2x)]=2^[10x] Thus, 10x-30=10x => -30=0 is a contradiction => the equation has no solution.

    5 Answers · Science & Mathematics · 11/03/2021

  2. ... - 3n - 1) P(1): 1² = 1/2(1)[6(1)² - 3(1) - 1] LHS = 1 RHS = 1 Proposition P(1) is confirmed...

    3 Answers · Science & Mathematics · 20/01/2021

  3. ... as x² = (2.3 - x)*1.19 * 10 ^-7. Then put all on the LHS . Then expand. Notice now that you have a quadratic of the form ...

    3 Answers · Science & Mathematics · 17/01/2021

  4. RHS = 2tan(2x)sin²(x) = tan(2x){1 - [1 - 2sin²(x)]} = tan(2x)[1 - cos(2x)] = tan(2x) - tan(2x)cos(2x) = tan(2x) - sin(2x) = LHS

    2 Answers · Science & Mathematics · 15/01/2021

  5. ...3 terms to evaluate: (3-6)x + (4-6)x + (5-6)x = 72. Simply the LHS . What does it give? [ = -6x ] Then, solving for x ...

    2 Answers · Science & Mathematics · 11/11/2020

  6. Yep, ordinary algebra.  Divide both sides by the same value and the equation remains balanced.  The LHS still = the RHS.  So divide away and enjoy.

    3 Answers · Science & Mathematics · 19/08/2020

  7. ... side  = (2^x)² - (2^x) * 2 + 1........... Put 2^x  =  a => LHS   =   a² - 2 a + 1   => LHS   =  ( a - 1 )² Hence...

    2 Answers · Science & Mathematics · 17/04/2021

  8. RHS = cot²(x)[csc³(x) - csc(x)] = cot²(x)csc(x)[csc²(x) - 1] = cot²(x)csc(x)cot²(x) = cot⁴(x)csc(x) = LHS

    1 Answers · Science & Mathematics · 09/05/2020

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