# Solve for "X" in this equation?

Solve for the equation below:

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

one week of not going to school has deteriorated my brain; therefore, I forgot how to do this.

But I think it requires a process using " Log" or " ln "

Thanks

### 8 Answers

- Anonymous1 decade agoFavorite Answer
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.

- ComoLv 71 decade ago
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.

- JacquelineLv 44 years ago
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- 1 decade ago
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

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- Anonymous1 decade ago
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.

- Pi R SquaredLv 71 decade ago
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!! :-)

- 1 decade ago
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!

- An ESL LearnerLv 71 decade ago
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]