Solve for "X" in this equation?

GIVEN

7 ^ (- 4x) = 2 ^ (1+3x)

and taking the natural logarithm of both sides,

-4x(ln 7) = (1 + 3x)(ln 2)

Noting that

ln 7 = 1.94591 and ln 2 = 0.693147, the above becomes

-4x(1.94591) = (1 + 3x)(0.693147)

-4x(2.807355) = 1 + 3x

-11.22942x = 1 + 3x

14.22942x = -1

x = -0.070277

You can also use common logarithm instead of natural logarithm. The process remains the same and definitely, the answer should be the same as well. I am simply comfortable in using natural logarithm.

Hope this helps.

What is a quadratic equation? Quadratic equations can come in many forms: x2 + 19 = 0 4x2 – 2x = 0 3x2 – 2x + 10 = 0 However, they are all in terms of one variable with a degree of 2.The standard form is ax2 + bx + c = 0, but b or c may be 0 as in the first and second equations above. Four Methods for Solving Quadratic Equations Factoring Method. Graphing Method. Square Root Method. Quadratic Formula Method. Why so many? Some quadratics factor easily. Some are easier to graph. If the quadratic equation does not factor then you will not be able to find a solution by factoring or an exact solution on a graph. Then the square root method is used. A short cut for the square root method is the quadratic formula. So, you can see that they all have their uses; it just depends on the problem. The good news is that you have learned the first two previously. In this section we will use the 3rd and 4th techniques listed above, involving square roots. The Square Root Method Binomials in the Form ax2 – c = 0 or ax2 = c. Isolate the x2. Take the square root of both sides. Simplify the two solutions. Check by substituting the solutions into the equation.

A well presented question due to brackets being used.

(- 4x) log 7 = (1 + 3x) log 2
(- 4x) log 7 = log 2 + (3 log 2) x
(4 log 7 + 3 log 2) x = - log 2
x = - log 2 / (4 log 7 + 3 log 2)

Any log base may be used to obtain a numerical answer.

7^(-4x) = 2^(1 + 3x)
log[7^(-4x)] = log[2^(3x + 1)]
(-4x)log(7) = (3x + 1)log(2)
-4x = (3x + 1)log(2)/log(7)
-4x = (3x + 1)log_7(2)
-4x = (3x)log_7(2) + log_7(2)
(3x)log_7(2) + 4x = -log_7(2)
x[3log_7(2) + 4] = -log_7(2)
x = [-log_7(2)]/[3log_7(2) + 4]

7^(- 4x) = 2^(1+3x)
log (7^(- 4x)) = log (2^(1+3x)) Take log of both sides.
-4x log 7 = (1+3x) log 2 Bring exponents out using log rule.
-4x log 7 = log 2 + 3x log 2 Distribute log 2 * (1+3x)
-4x log 7 - 3x log 2 = log 2 Get x's on one side.
x (-4 log 7 - 3 log 2) = log 2 Factor
x = (log 2) / (-4 log 7 - 3 log 2) ANSWER

I think you're right.
Take the log of both sides:
-4x log 7 = (1+3x) log 2

Plug in the values of log 7 and log 2, and then solve the linear equation.

Hi,

7 ^ (- 4x) = 2 ^ (1+3x)

-4x log 7 = (1+3x) log 2

-4x log 7 = log 2 + 3x log 2

- log 2 = 3x log 2 + 4x log 7

- log 2 = x(3 log 2 + 4 log 7)

- log 2
------------------------ = x
3 log 2 + 4 log 7

x = -.0703 <==ANSWER

I hope that helps!! :-)

ln(7^(-4x))=ln(2^(1+3x))
(-4x)ln(7)=(1+3x)ln(2)
ln(2)=x(-4ln(7)-3ln(2))
x=ln(2)/(-4ln(7)-3ln(2))
x=-0.0703

When you take the logarithm of an expression, you can multiply the logarithm by the exponent in the expression. Hope this helps!