Exponential, solve for x

[K_Swiss]

Solve for x.

\(\displaystyle e^{0.113x} + 4.72 = 7.031 - x\)

\(\displaystyle e^{0.113x} + x = 7.-31 - 4.72\)

\(\displaystyle e^{0.113x} + x = 2.314\)

\(\displaystyle ln(e^{0.113x}) + ln(x) = ln(2.314)\)

\(\displaystyle 0.113x + lnx = ln2.314\)

\(\displaystyle x(0.113 + 0) = ln2.314\)

\(\displaystyle x = \frac{ln2.314}{0.113}\)

\(\displaystyle x = 7.42\) ... and my answer is wrong since the book answer is \(\displaystyle 1.1697\). What did I do wrong?

 

[galactus]

Solve for x.

\(\displaystyle e^{0.113x} + 4.72 = 7.031 - x\)

\(\displaystyle e^{0.113x} + x = 7.-31 - 4.72\)

\(\displaystyle e^{0.113x} + x = 2.314\)

\(\displaystyle \underbrace{ln(e^{0.113x}) + ln(x)}_{\text{Here....illegal move}} = ln(2.314)\)

Should be \(\displaystyle ln(e^{.113x}+x)=ln(2.311)\). You can't take the ln of them separately like that.

 

[fasteddie65]

Solve for x.

\(\displaystyle e^{0.113x} + 4.72 = 7.031 - x\)

\(\displaystyle e^{0.113x} + x = 7.-31 - 4.72\)

\(\displaystyle e^{0.113x} + x = 2.314\)

\(\displaystyle ln(e^{0.113x}) + ln(x) = ln(2.314)\)

\(\displaystyle 0.113x + lnx = ln2.314\)

\(\displaystyle x(0.113 + 0) = ln2.314\)

\(\displaystyle x = \frac{ln2.314}{0.113}\)

\(\displaystyle x = 7.42\) ... and my answer is wrong since the book answer is \(\displaystyle 1.1697\). What did I do wrong?

The basic difficulty with a problem like this is that it cannot be solved with algebra alone. The expression e^(0.113x) is a transcendental function, while x is an algebraic one. They are not compatible, so the best method is to use a graphing calculator. Put the left-hand side as Y1 and the right-hand side as Y2, and find the intersection point(s). I did, and found the same answer as the book.

 

[galactus]

Are you familiar with Newton's method?. It converges quickly in this case if you use an initial guess of 1.5.

 

[jonah]

Slight typo.
7.031 - 4.72 = 2.311 and not 2.314
Used MAPLE to see what a CAS might do with this.
Turned out that the Lambert W function was needed on this one (not that I'm in any way familiar with it).
See http://en.wikipedia.org/wiki/Lambert's_W for more info.