Solve for x using logs:

1. 3^(2x+1) = 6
Note: 3 to the power of (2x+1) = 6

2. 3 * 4^(x-1) = 5 * 3^(x+1)

asked Feb 26, 2013 in CALCULUS

32x+1 =6

Apply log each side.

log32x+1 =log6

Recall: logarithm formula log am   =m log(a)

(2x+1)log3 = log6

(2x)log3  + log3   =log6

Subtract log3 from each side.

2xlog3   =log6 - log3

2xlog3=log6/log3

Recall: logarithm formula log a - logb=log(a/b)

2xlog3=log(6/3)

2xlog3=log2

Divide each side by 2log3

x = (1/2)[log2/log3].

answered Mar 4, 2013

2).

The equation is 3 * 4x - 1 = 5 * 3x + 1 .

Apply natural log on both sides.

ln (3 * 4x - 1 ) = ln (5 * 3x + 1)

Apply product rule : ln (uv) = ln u + ln v.

ln 3 + ln (4x - 1) = ln 5 + ln (3x + 1)

Apply power rule : ln am   =m ln(a)

ln 3 + (x - 1) ln 4 = ln 5 + (x + 1)ln 3

(x - 1) ln 4 - (x + 1)ln 3 = ln 5 - ln 3

x ln 4 - ln 4 - x ln 3 - ln 3 = ln 5 - ln 3

x ln 4 - x ln 3 = ln 5 + ln 4

x (ln 4 - ln 3) = ln (5*4)

x (ln 4 - ln 3) = ln (20) / (ln 4 - ln 3)

x = 2.9957 / (ln 4 - ln 3)

x = 2.9957 / 0.2877

⇒ x = 10.41.

The solution is x = 10.41.

answered Jul 30, 2014