Additional Mathematics 0606 / Logarithmic and exponential functions / 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← 1.14: Logarithms in terms of p and q; exponential equation1.3: Logarithmic equations with bases 3, 9 and 4 →
