Additional Mathematics 0606 / Logarithms and Exponents / 2.62.6: Exponential equation; lg equation with two factors0606/21/O/N/19 — Question 3 · 8 marks · Exponentials, Log laws, EquationsMark as done · Save for later · Show all solutions(a)Solve e2x+1=3e4−3xe^{2x + 1} = 3e^{4 - 3x}e2x+1=3e4−3x.[3]▸ Answer(b)Solve lg(y−6)+lg(y+15)=2\lg(y - 6) + \lg(y + 15) = 2lg(y−6)+lg(y+15)=2.[5]▸ Answer← 2.5: Exponential equations in base 3 and base e2.7: Write as a single logarithm →
