Mr Daniels Maths
Solving Equations with indices

Set 1

Set 2

Set 3

Q1) Solve \(8^{5 x -3} = 8^{7}\). x= [ 2]

Q1) Solve \(5^{3x + 19} = 125^{6}\). x= [ -\(\frac{1}{3}\)]

Q1) Solve \(5^{4x + 10} = 3125^{3x -8}\) . x= [ 4\(\frac{6}{11}\)]

Q2) Solve \(4^{3 x -10} = 4^{5}\). x= [ 5]

Q2) Solve \(3^{4x + 5} = 81^{-2}\). x= [ -3\(\frac{1}{4}\)]

Q2) Solve \(5^{10x + 6} = 625^{6x -7}\) . x= [ 2\(\frac{3}{7}\)]

Q3) Solve \(4^{5 x +7} = 4^{2}\). x= [ -1]

Q3) Solve \(2^{6x + 9} = 16^{3}\). x= [ \(\frac{1}{2}\)]

Q3) Solve \(4^{9x + 5} = 1024^{9x + 3}\) . x= [ -\(\frac{5}{18}\)]

Q4) Solve \(7^{2 x -7} = 7^{7}\). x= [ 7]

Q4) Solve \(5^{9x + 7} = 25^{3}\). x= [ -\(\frac{1}{9}\)]

Q4) Solve \(2^{9x + 7} = 8^{7x + 9}\) . x= [ -1\(\frac{2}{3}\)]

Q5) Solve \(8^{7 x -6} = 8^{8}\). x= [ 2]

Q5) Solve \(2^{6x + 5} = 8^{-5}\). x= [ -3\(\frac{1}{3}\)]

Q5) Solve \(2^{2x + 3} = 4^{8x -6}\) . x= [ 1\(\frac{1}{14}\)]

Q6) Solve \(6^{2 x +8} = 6^{2}\). x= [ -3]

Q6) Solve \(3^{4x + 13} = 81^{-2}\). x= [ -5\(\frac{1}{4}\)]

Q6) Solve \(2^{9x + 3} = 32^{6x + 5}\) . x= [ -1\(\frac{1}{21}\)]

Q7) Solve \(7^{5 x -8} = 7^{2}\). x= [ 2]

Q7) Solve \(2^{5x + 17} = 8^{4}\). x= [ -1]

Q7) Solve \(2^{6x + 3} = 4^{5x + 10}\) . x= [ -4\(\frac{1}{4}\)]

Q8) Solve \(9^{10 x -3} = 9^{7}\). x= [ 1]

Q8) Solve \(5^{10x + 12} = 625^{-6}\). x= [ -3\(\frac{3}{5}\)]

Q8) Solve \(2^{10x + 3} = 4^{7x -9}\) . x= [ 5\(\frac{1}{4}\)]

Q9) Solve \(5^{3 x +3} = 5^{9}\). x= [ 2]

Q9) Solve \(4^{8x + 6} = 256^{-10}\). x= [ -5\(\frac{3}{4}\)]

Q9) Solve \(3^{4x + 4} = 27^{5x + 3}\) . x= [ -\(\frac{5}{11}\)]

Q10) Solve \(5^{2 x +3} = 5^{7}\). x= [ 2]

Q10) Solve \(5^{8x + 2} = 125^{6}\). x= [ 2]

Q10) Solve \(3^{4x + 5} = 27^{5x + 2}\) . x= [ -\(\frac{1}{11}\)]