Exercise 74 Page 483

We want to solve the exponential equation. When bases are not the same, we can solve it by taking the logarithm of each side of the equation. m=n ⇔ log m = log n Note that in order to take their logarithms, both m and n must be positive numbers. Let's now solve our equation.

25^(2x+1)=144

log_()(LHS)=log_()(RHS)

log 25^(2x+1)= log 144

log_()(a^m)= m* log_()(a)

(2x+1) log 25 = log 144

DivEqn

.LHS /log 25.=.RHS /log 25.

2x+1=log 144/log 25

UseCalc

Use a calculator

2x+1 ≈ 1.544

SubEqn

LHS-1=RHS-1

2x≈ 0.544

DivEqn

.LHS /2.=.RHS /2.

x≈ 0.272