Going crazy with logarithims!!!!!!!!!!!?

Question

Can anyone please help with the following problem? We are studing solving exponential equations.

2^(3x+1) = 3^9x-2)

I get that far, but am having trouble with isolating x

Answer 1

To start you are going to take the log of both sides. You will end up using the power rule of logs that says

log (x^y) is equal to y*log (x)

So basically if there is a power and you take the log, you can move the power out in front to multiply.

2^(3x+1) = 3^(9x-2)
log[2^(3x+1)]= log[3^(9x-2)]
(3x+1)log(2) = (9x-2)log(3)

Each log now is just a constant, so distribute.

log(2)*3x +log(2)*1 = log(3)*9x - log(3)*2

Which would look better this way

(3x)log(2) + log(2) = (9x)log(3) - (2)log(3)

Now get the variables on one side and the constants (anything w/o and x) on the other.

(3x)log(2) - (9x)log(3) = -log(2) - (2)log(3)

Factor out an x on the left side.

x[(3)log(2) - (9)log(3)] = -log(2) - (2)log(3)

You could do some elementary simplifying at this step, but I'm going to divide by all the stuff inside the brackets to get x by itself.

x = [-log(2) - (2)log(3)]/[(3)log(2) - (9)log(3)]

Stick all that in your calculator and you're done!

Answer 2

take the natural log of both sides:

ln 2^(3x+1) = ln 3^(9x-2)

Use the property ln a^b = b ln a:

(3x+1) ln 2 = (9x-2) ln 3

Isolate all terms with x on one side:
(3 ln 2 - 9 ln 3)x = -ln 2 - 2 ln 3

and divide to obtain x. You'll need a calculator for a numerical value.

Answer 3

You need to take the logarithm (best to use base 10 or base e) of both sides. Then use the power rule for logarithms:

ln2^(3x + 1) = ln3^(9x - 2)

(3x + 1)ln2 = (9x - 2)ln3

(3ln2)x + ln2 = (9ln3)x - 2ln3

Now collect the two terms involving x on the left, the two constant terms on the right, and factor x on the left:

(3ln2)x -(9ln3)x = -ln2 - 2ln3

x(3ln2 - 9ln3) = -(ln2 + 2ln3)

x = -(ln2 + 2ln3)/(3ln2 - 9ln3)

Use a calculator to compute the right side to find the value of x!

Answer 4

If you apply logs to both sides you end up with:

(3x + 1) * log 2 = (9x - 1) * log 3

And solve from here. First order equation in x.

Answer 5

Assuming you meant: 2^(3x+1) = 3^(9x-2)

(2^3x)(2) = (3^9x)(3^-2) = (3^9x)(1/9)
18 = (3^9x)/(2^3x)
ln18 = ln (3^9x)/(2^3x) = ln(3^9x) - ln(2^3x) = 9x*ln3 - 3x*ln2
ln18 = x(9*ln3 - 3*ln2)
x = ln18/(9*ln3 - 3*ln2)

Answer 6

2^(3x+1) = 3^(9x-2)

ln 2^(3x+1) = ln 3^(9x-2)

(3x + 1)ln 2 = (9x - 2)ln 3

3x*ln 2 + ln 2 = 9x*ln 3 - 2*ln 3

9x*ln 3 - 3x*ln 2 = ln 2 + 2*ln 3

x(9*ln 3 - 3*ln 2) = ln 2 + 2*ln 3

x = (ln 2 + 2*ln 3) / (9*ln 3 - 3*ln 2)

x = 0.37

Answer 7

2^(3x+1) = 3^(9x-2) take ln of each side
(3x+1)ln(2)=(9x-2)ln(3)
x(3ln(2)-9ln(3))=-2ln(3)-ln(2)
x=-(2ln(3)+ln(2))/(-(9ln(3)-2ln(3)))=.37018