Using Properties of Logarithms

When bases are not the same, we can solve an exponential equation by taking logarithms on each side of the equation.

m = n \Leftrightarrow lo g (m) = lo g (n)

Note that in order to take their logarithms, both

m

and

n

must be positive numbers. Let's now solve our equation by using the properties of logarithms.

2 5^{2 x + 1} = 144

lo g (LHS) = lo g (RHS)

lo g (2 5^{2 x + 1}) = lo g (144)

lo g (a^{m}) = m \cdot lo g (a)

(2 x + 1) lo g (25) = lo g (144)

LHS / lo g (25) = RHS / lo g (25)

2 x + 1 = \frac{lo g ( 1 4 4 )}{lo g ( 2 5 )}

UseCalcUse a calculator

2 x + 1 = 1.543959311 \dots

LHS - 1 = RHS - 1

2 x = 0.543959311 \dots

LHS / 2 = RHS / 2

x = 0.2719796553 \dots

RoundDecRound to

x \approx 0.2720

Using Properties of Logarithms 1.2 - Solution