Logarithmic Equation

[0313phd]

PROBLEM: Log10x + log10(x-21) = 2
I converted to: log10 x(x-21)= 100
log 10 xsquared -21x=100
It didn't seem to factor, so I tried the quadratic equation. I got approximately 24 as the answer, but the book gives 25 as the answer. What did I do wrong?

 

[JeffM]

PROBLEM: Log10x + log10(x-21) = 2
I converted to: log10 x(x-21)= 100 Is this a typo? How did 2 become 100? Let's take it one step at a time.
log 10 xsquared -21x=100
It didn't seem to factor, so I tried the quadratic equation. I got approximately 24 as the answer, but the book gives 25 as the answer. What did I do wrong?

 

[0313phd]

No, its not a typo. Originally, I had log10x + log 10(x-21)=2. I got 100 because 10 is the base, to the second power or 2, equals 100. So, I converted to: log10 x(x-21)= 100
log 10 xsquared -21x=100
It didn't seem to factor, so I tried the quadratic equation. I got approximately 24 as the answer, but the book gives 25 as the answer. What did I do wrong?

 

[soroban]

Hello, 0313phd!

\(\displaystyle \log_{10}x + \log_{10}(x-21) \:=\: 2\)

\(\displaystyle \text{I converted to: }\:\log_{10}x(x-21)\:=\:100\) . . . . no

\(\displaystyle \text{You have: }\:\log_{10}x(x-21) \:=\:2\)

\(\displaystyle \text{which becomes: }\:x(x-21) \:=\:100 \quad\Rightarrow\quad x^2 - 21x - 100 \:=\:0\)

. . . . \(\displaystyle (x - 25)(x + 4) \:=\:0 \quad\Rightarrow\quad x \:=\:25,\:\rlap{//}\text{-}4\)

\(\displaystyle \text{Therefore: }\:x \,=\,25\)

 

[JeffM]

Soroban gave you the right answer of course, but he did not explain two critical steps.

You properly SIMPLIFIED log[sub:16o7jmbj]10[/sub:16o7jmbj](x) + log[sub:16o7jmbj]10[/sub:16o7jmbj](x - 21) to log[sub:16o7jmbj]10[/sub:16o7jmbj](x[sup:16o7jmbj]2[/sup:16o7jmbj] - 21x).

You cannot, however, say 2 = 100 just because 2 = log[sub:16o7jmbj]10[/sub:16o7jmbj](100). 2 NEVER equals 100.

So, the equation you SHOULD have come up with was log[sub:16o7jmbj]10[/sub:16o7jmbj](x[sup:16o7jmbj]2[/sup:16o7jmbj] - 21x) = 2 = log[sub:16o7jmbj]10[/sub:16o7jmbj](100).

Now, and this is a critical step. If log[sub:16o7jmbj]b[/sub:16o7jmbj](z) = log[sub:16o7jmbj]b[/sub:16o7jmbj](y), then z = y. Note it is NOT generally true that if log[sub:16o7jmbj]c[/sub:16o7jmbj](z) = log[sub:16o7jmbj]d[/sub:16o7jmbj](y), then y = z. The bases must be the same.

So, from log[sub:16o7jmbj]10[/sub:16o7jmbj](x[sup:16o7jmbj]2[/sup:16o7jmbj] - 21x) = log[sub:16o7jmbj]10[/sub:16o7jmbj](100), you can eliminate the log function from both sides of the equation and get x[sup:16o7jmbj]2[/sup:16o7jmbj] - 21x = 100.

Now you can factor or complete the square or factor to your heart's content. And you can check your work.

log[sub:16o7jmbj]10[/sub:16o7jmbj](25) + log[sub:16o7jmbj]10[/sub:16o7jmbj](25 - 21) = log[sub:16o7jmbj]10[/sub:16o7jmbj](25) + log[sub:16o7jmbj]10[/sub:16o7jmbj](4) = log[sub:16o7jmbj]10[/sub:16o7jmbj](25 * 4) = log[sub:16o7jmbj]10[/sub:16o7jmbj](100) = 2.

All clear now?

PS Soroban also did not explain why x = - 4 is not a valid answer. Do you know why?

 

[0313phd]

Yes, thank you. I am all clear now. Negative 4 is not a valid response because it is not in the domain for logarithmic functions. Did I explain that correctly? THANKS for all of your help. 0313

 

[JeffM]

Yes, thank you. You're welcome.
I am all clear now. GREAT
Negative 4 is not a valid response because it is not in the domain for logarithmic functions. Did I explain that correctly? Absolutely.

 

[lookagain]

Negative 4 is not a valid response because it is not in the domain f
or logarithmic functions. Did I explain that correctly?

No,

while -4 is not in the domain of either of the two log functions in this equation,
it is possible for certain logarithmic functions to have negative numbers as part of their domains.

0313phd,

it is not the fact that the value of x being negative would make it automatically
not a solution of \(\displaystyle \text{this}\) logarithmic equation.


For instance,

\(\displaystyle the \ equation \ \log_2(x + 12) \ = \ 3\)

\(\displaystyle \ has \ the \ solution \ of \ x \ = \ -4.\)

 

[0313phd]

Look again,

Is that because -4 is the only possible solution? 0313

 

[lookagain]

Look again,

Is that because -4 is the only possible solution? 0313

x = -4 is the only candidate, but it must be tested in one
of the proper equations, such as the original one.

And it satisfies one of those proper equations, so it is the only solution.

However, if x = -4 (the only candidate) had not checked, then
the answer would be "no solution."