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solve for x 49^(3x+1)=7^((x^2)+16x+18)
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Notice that 49 = 7^2, so the LHS 49^(3x+1) = 7^(6x+2). Now LHS = RHS if and only if the exponents are equal; i.e, 6x + 2 = x^2 + 16 x + 18 or x^2 + 10x + 16 = 0. This is straight-forward to solve. I recommend NOT using Wolfram, as you'll miss the insight of figuring out the exponents. And you're not going to have Wolfram in your exam either! ;-)
(3x+1)ln(49)=(x^2+16x+18)ln(7) since ln(49) and ln(7) are constants get all the variables on one side n solve.
(3x+1)ln(49)=(x^2+16x+18)ln(7) since ln(7)^2=ln(49) using this 2ln(7)=ln(49) 2(3x+1)ln(7)=(x^2+16x+18)ln(7) the ln(7) cancel leaving 6x+2=x^2+16x+18 0=x^2+10x+16 (x+8)(x+2)=0 x=-8 x=-2
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