Quick Calculus Question?

Question

Okay, I am kind of clueless right now..please help and explain if you can.. Thank you

Solve for x: 2^1-x = 3^x

The answer is ln 2 / ln 6, but I have no clue how to get that.

Answer 1

2^(1-x) = 3^x

ln of both sides:
ln[2^(1-x)] = ln[3^x]

Exponent rule for logs:
(1-x) ln(2) = x ln(3)

Distribute ln(2):
ln(2) - x ln(2) = x ln(3)

Add xln(2) to both sides:
ln(2) = x ln(3) + x ln(2)

Take out an x on the right side:
ln(2) = x[ln(3) + ln(2)]

Mult./add. rule for logs:
ln(2) = x[ln(3*2)]
ln(2) = x ln(6)

Divide both sides by ln(2):
x = ln(2) / ln(6)

Answer 2

2^(1 - x) = 3^x
(1 - x) ln 2 = x ln 3
(1 - x) / x = ln 3 / ln 2
1 / x - 1 = ln 3 / ln 2
1 / x = ln 3/ ln 2 + 1
1 / x = ( ln 3 + ln 2) / ln 2
x = ln 2 / ln 6

Answer 3

Use the following property of logarithms:

log(a^b) = b*log(a).

So (1-x)ln(2) = xln(3)

Now use straightforward algebra to solve for x (and remember that ln(ab) = ln(a) + ln(b)).

Math Rules!

Answer 4

2^(1-x) = 3^x;
3 = 2^(1-x)/x;
ln 3 = [(1-x) / x ] ln 2;
(1-x)/x = ln 3/ln 2;
1/x - 1 = ln 3/ln 2;
1/x = (ln 3/ln 2 )+ 1;
1/x = (ln 3 + ln 2)/ln 2;
1/x = ln (3)(2)/ln 2; (since ln a + ln b = ln (ab))
1/x = ln 6 /ln 2;
x = ln 2/ ln 6

Answer 5

2/2^(x) = 3^x
.
2=3^x 2^x
.
2=6 ^x
.
ln2 =ln 6^x
ln2 =x ln6
ln/ln6 =x
.

Answer 6

(1-x)log2 = xlog3
1log2 - xlog2 = xlog3
log2 = xlog3 + xlog2
log2 = x(log3+log2)
log2/(log3+log2) = x

Answer 7

x= 0.5