I'm supposed to solve using logrithims

Hello Susie,

(7)^{3/x} = 5^{(2 - x) } original equation

log_{7}(7)^{3/x} = log_{7}(5)^{(2 - x) } Take log (base 7) of each side.

(3/x)log_{7}7 = (2 - x)log_{7}5 Use log_{a}u^{n} = n log_{a}u

(3/x) * 1 = (2 - x)log_{7}5 log_{a}a = 1 because a^{1} = a

3/x = (2 - x)(log5/log7) change of base formula log_{a}x = log x/log a

3/x = (2 - x)(0.6989/0.8451) use calculator to find the values of log5 and log7

3/x = (2 - x)(0.827)

3 = (2 - x)(0.827)(x) Multiply by x on both sides

3/0.827 = (2 - x)(x) divide by 0.827 on both sides

3.628 = 2x - x^{2 }distribute x on right side

Finally, it is a quadratic equation

x^{2} - 2x + 3.628 = 0

Here a = 1, b = -2 and c = 3.628

Use the quadratic formula and solve for x. You'll get two values of x as an answer. Just plug in the values of a, b, and c in the formula.

x = (-b ± √(b^{2} - 4ac))/2a

Good luck. If you need any help you can ask.

## Comments

Roman's attack is more elegant, in that it avoids use of the quadratic formula. Completing the square is easier. The end result is the same. Seeing more than one attack of a problem can make a problem clearer.