Logarithms

use logarithms to solve equations in

which the unknown appears in indices

Since log_ax only exists for x > 0 and for a > 0 and a≠1, it is essential when solving equations involving logs that all roots are checked in the original equation

2 log₈(x+2)=log₈(2x+19)

log₈(x+2)²=log₈(2x+19)

(x+2)² = 2x+19

x²+4x+4=2x+19

x²+2x-15=0

(x-3)(x+5)=0

x=3 or x=-5

So, x=3

4  log_x2−log_x4=2

log_x2⁴−log_x2²=2

log_x2⁴⁻²=2

log_x2²=2

log_x4=2

x²=4 -------> x=2 or x=-2

So, x = 2

Solve, giving your answers correct to 3 significant figures.

5^2x+1=7

log  5^2x+1=log  7

(2x+1)log  5=log  7

2x  log  5+log  5=log  7

x=log  7−log  5/2  log  5

x=0.105

Solve, giving your answers correct to 3 significant figures.

3^2x = 4^x+5

log  3^2x = log  4^x+5

2x  log  3 = (x+5)log  4

2x  log  3 = x  log  4+5  log  4

x(2  log  3−log  4)=5  log  4

x=5  log  4/2  log  3−log  4

x=8.55