These two are supposed lớn be equivalent because of the properties of logarithms, but the domains of \$ln(x^2)\$ và \$2ln x\$ seem different lớn me. For example, if I substitute \$x=-1\$ into the first, I get 0. But in the second, I get a non-real answer.

Why is this? The domains of these functions, when graphed, seem different as well. But everything I"ve been taught so far, & most of the things I can find on the web, bởi vì not explain this inconsistency. Typically, I consider the property \$ln(a^b) = b ln a\$ to lớn be true...

Bạn đang xem: How do you solve lnx=2?

I am only a senior in high school, currently in Calculus I, but hopefully the explanation won"t be too outside of the realm of my understanding. But even if it is, I would still lượt thích to know the answer.

logarithms
tóm tắt
Cite
Follow
asked Apr 6, 2011 at 14:44

ReidReid
\$endgroup\$
1
địa chỉ a phản hồi |

Sorted by: Reset to mặc định
Highest score (default) Date modified (newest first) Date created (oldest first)
29
\$egingroup\$
You are right. In the reals, you can define \$ln(x^2)\$ for all of \$ omanhords.combbR\$ except 0, while \$2ln x\$ is only defined for \$x gt 0\$. Where they are both defined, they agree. The proper way lớn write the equality is \$ln (x^2)=2ln |x|\$, in which case the domains agree. People get sloppy sometimes.

tóm tắt
Cite
Follow
answered Apr 6, 2011 at 14:51

Ross MillikanRoss Millikan
\$endgroup\$
5
địa chỉ a bình luận |
12
\$egingroup\$
When the identity is stated as \$ln(a^b)=bln(a)\$, it is assumed that \$a\$ is positive. Thus, regardless of whether it is possible to extend \$ln(a^b)\$ lớn a larger range of \$a\$ values for special choices of \$b\$, this is a valid identity with this domain restriction.

If the case where \$b\$ is an even integer were singled out, then it would be good khổng lồ point out that the identity can be generalized lớn \$ln(a^b)=bln(|a|)\$ for all nonzero \$a\$. But the more general (domain restricted) identity allows \$b\$ to lớn be any real number, including not only odd integers, but fractions, & even irrational numbers. In most cases, \$ln(a^b)\$ wouldn"t be defined in an ordinary (real) sense when \$a\$ is negative.

To properly define a function, its tên miền should be specified. Things aren"t always done properly. For example, a typical question used in precalculus classes gives a formula defining a function, like \$displaystylef(x)=fracsqrtx+4x-2\$ and asks the student to "find the domain". The idea is that you"re supposed to figure out all possible numbers you can plug into the formula such that the result is defined as a real number. This is a good exercise because it can help lớn build familiarity with the functions & test algebra skills, but it can be misleading. The tên miền can depend on context, & whoever is defining the function can decide. E.g., if \$f\$ is the function defined on \$(0,infty)\$ by \$f(x)=ln(x^2)\$, then I have stipulated the domain to be \$(0,infty)\$. The identity \$f(x)=2ln(x)\$ is valid on this domain.

Nonetheless, I absolutely agree with you và Ross that if you"re just starting with \$ln(x^2)\$ with no context khổng lồ imply that \$x\$ is positive, then the negative case has lớn be taken into account, và thus the more general identity \$ln(x^2)=2ln(|x|)\$ would be required.

Similarly, in the identity \$ln(ab)=ln(a)+ln(b)\$, it is assumed that \$a\$ và \$b\$ are positive. The left-hand side would also be defined when \$a\$ & \$b\$ are both negative, & in general you would have \$ln(ab)=ln(|a|)+ln(|b|)\$ whenever \$a\$ & \$b\$ have the same sign. The fact that \$ln(ab)\$ can be defined while \$ln(a)\$ & \$ln(b)\$ are not leads khổng lồ so-called "extraneous solutions" in precalculus problems on solving logarithmic equations, if one is not careful about domain restrictions.

Xem thêm: Điểm Chuẩn Đại Học Bách Khoa Năm 2021, Điểm Chuẩn Đại Học Bách Khoa

Example Problem: Solve the equation \$ln(x)+ln(x+1)=-10\$.

Solution: Using the identity \$ln(ab)=ln(a)+ln(b)\$, the equation becomes \$ln(x^2+x)=-10\$. Exponentiating yields \$x^2+x=frac1e^10\$. Solving the quadratic equation, \$x=-frac12pmfrac12sqrt1+4/e^10\$. But wait, one of these doesn"t work in the original equation, because \$-frac12-frac12sqrt1+4/e^10\$ is negative. This "solution" is a solution to lớn \$ln(x^2+x)=-10\$, but not lớn the original equation, because \$ln(x^2+x)\$ has larger domain than \$ln(x)+ln(x+1)\$.

(Also, you can see that there should only be one solution to the original equation, because \$ln(x)+ln(x+1)\$ is always increasing where it is defined.)