5 Log problems

[jonboy]

Hey guys/girls my teacher been pretty quick on this topic of logarithms so I would like you to check these 5 probs -

The direction are: Solve for x and approximate to the nearest hundredth if necessary.

1) \(\displaystyle \L \;e^{-\,4x}\,-\,22\,=\,56\)

\(\displaystyle \L e^{\,-\,4x}\,=\,78\)

Take log of both sides and rearrange:\(\displaystyle \L \;\,-\,4xlog_{e}\,=\,log\,78\)

\(\displaystyle \L \;-\,4x\,*\,.43\,=\,1.89\)

...............\(\displaystyle \L x\,\approx\,-\,1.01\)


2) \(\displaystyle \L \;3\,ln\,x\,=\,ln\,8\)

\(\displaystyle \L 3\,e^{ln}\,x\,=\,e^{ln}\,8\)

Which leads me with:\(\displaystyle \L \;3x\,=\,8\,\to\,x\,=\,\frac{8}{3}\)

3)\(\displaystyle \L \;\,ln\,x\,+\,ln\,(x\,+\,1)\,=\,ln\,2\)

\(\displaystyle \L ln\,x\,+\,ln\,x\,=\,ln\,2\)

...............\(\displaystyle \;ln\,(x\,+\,x)\,=\,ln\,2\)

\(\displaystyle \L \;e^{ln}(x\,+\,x)\,=\,e^{ln}\,2\)

\(\displaystyle \L \;2x\,=\,2\)

\(\displaystyle \L x\,=\,1\;\;\) I know this is wrong but where and why did I go wrong?

4)\(\displaystyle \L \;2\,ln\,(x\,+\,2)\,=\,1\)

\(\displaystyle \L \;2\,ln\,2x\,=\,1\)

...............\(\displaystyle \L \;2\,e^{ln}\,2x\,=\,e\)

\(\displaystyle \L \;x\,=\,\frac{e}{4}\)

5)\(\displaystyle \L \;3\,log\,x\,+\,7\,=\,3\)

\(\displaystyle \L \;3\,log\,x\,=\,-\,4\;\,\) What is the base of the log on the left?

\(\displaystyle \L \;3\,10^{log}\,x\,=10^{\,-\,4}\)

\(\displaystyle \L \;3x\,=\,\frac{1}{10^4}\)

...............\(\displaystyle \L \,x\,=\,\frac{1}{30\,,\,000}\)

Thank you soo much if you help me!

 

[tkhunny]

You have introduced rounding far too quickly and carelessly.

log(78) = 1.89209460
log(e) = 0.43449428
x = -1.08917721

NOW round to hundreths. It is a VERY different answer.

By the way, what does \(\displaystyle \L\;\log_{e}\) mean? I see that you should have intended to write log(e), but that is not what you have.

 

[tkhunny]

No, no, and no.

Again, what does this mean? \(\displaystyle \L\;e^{ln}\) It is a meaningless expression. Never write it again.

3*ln(x) = ln(8)

ln(x^3) = ln(8)

x^3 = 8

x = 2

I cannot tell what it is you were doing.

 

[tkhunny]

You must review your rules.

log(x+1) is NOT NOT NOT NOT equal to log(x).

This is going nowhere mighty fast. It is time for you to review your lesson materials. If you can blame it on your teacher, that is fine, but it still does you no good. You don't seem to be real close on these guys.

 

[stapel]

1) e^-4x - 22 = 56

Adding 22 to either side was fine, but the rest is wrong. I will guess that, by "log_e( )", you mean "I took the natural log of both side, and forgot the argument on the one side". But forgetting function arguments will lead to problems, as the earlier replies suggest. Instead, do each step clearly:

. . . . .e^-4x - 22 = 56

. . . . .e^-4x = 78

. . . . .ln(e^-4x) = ln(78)

. . . . .-4x ln(e) = ln(78)

Since ln(e) = 1 (because e¹ = e), then you have:

. . . . .-4x = ln(78)

Complete the solution. Only after you have "x=" should you pull out the calculator and do any approximations.

2) I'm afraid I can't figure out what you might mean by "e^{ln( )}"...? Instead, try isolating the x-term, and then raising both sides as powers on e. Since e^ln(f(x)) = f(x), you can then solve for "x=", at which point you can approximate.

3) If f(1) = 0, then does f(2) = 0 as well, "since" f(2) = f(1 + 1) "=" f(1) + f(1)...?

4) How did you get that x + 2 equalled 2x...?

5) If your textbook did not define its terms, then I'm afraid there is no way to tell what the base is meant to be. Sorry.

You mention that your instructor has covered the material very quickly. If you are feeling unprepared, you might want first to study up on logs, log rules, exponentials, solving log equations, and solving exponential equations, before proceeding any further.

Eliz.

 

[jonboy]

No, no, and no.

ln(x^3) = ln(8) How did you get rid of the ln?

x^3 = 8

 

[pka]

ln(x^3) = ln(8) How did you get rid of the ln?
x^3 = 8

On the domain of ln(x) the function is one-to-one.
If ln(a)=ln(b) then a=b.

 

[jonboy]

Ok guys I understand these problem now.

Just check me on the last one:

\(\displaystyle \L \;3\,log\,x\,+\,7\,=\,3\)

Subtract 7 and divide by 3:\(\displaystyle \L \;log\,x\,=\,\frac{\,-\,4}{3}\)


Raise both to the power of 10:\(\displaystyle \L \;x\,=\,10^{\frac{\,-\,4}{3}}\,=\,\frac{1}{10^{\frac{4}{3}}}\,=\,\frac{1}{10\sqrt[3]{10}}\)

Correct?

 

[Denis]

1) \(\displaystyle \L \;e^{-\,4x}\,-\,22\,=\,56\)

Here's my way (nothing wrong with yours!):
1 / e^(4x) = 78
e^(4x) = 1/78
4x = log(1/78) / log(e)
x = log(1/78) / 4
x = -1.0891772...

 

[tkhunny]

No, no, and no.

ln(x^3) = ln(8) How did you get rid of the ln?

x^3 = 8

Read pka's response very carefully. From your initial attempts, I expect you have concluded that I divided by "ln", This would be a very, very bad conclusion.

Also, I didn't just get rid of it. I still have to remember what it was. The Domain remains important, no matter what you do.

x^3 can take x < 0

ln(x^3) cannot take x < 0

Why?