Question

Write the expression as a single logarithm.

log_340 – log_310

Answer 1

There are rules you know!

Rules for log that everyone with log probs should know:

1. log_a(x) + log_a(y) = log_a(xy)
2. log_a(x) - log_a(y) = log_a(x/y)
3. log_a(x^k) = k*log_a(x)
4. log_a(x)=log_b(x)/log_b(a)

@Hero wrote your expression in the way you should have ;)

Just apply rule nr...?

Answer 2

Uh, loga4?

Answer 3

I'm not stupid, I just don't have time to learn this stuff considering my work is due in about 8 hours.

Answer 4

I know which rule it is. I read it when ZeHanz posted it. Is it Log_34?

Answer 5

Theres a 3 n there not an a? and x/y is 40/10 which simplifies to 4

Answer 6

I know you are not stupid and it was not my intention to imply that, sorry!
Not having much time is the reason why you should know about the rules.
You don't want to have to discover how it works everytime you get such a problem.

The answer is:\[\log_{3} 4\]

Answer 7

I did it wayyy up there ^ you just never read it :P

Answer 8

Solve the exponential equation. 1/16=64^4x-3. What is the rule for that?

Answer 9

(1/16)=64^(4x-3)

Answer 10

I did read it, and just wanted to confirm you were right...
64 and 1/16 are both powers of 2, it could be helpful to write these numbers that way:\[\frac{ 1 }{ 16 }=\frac{ 1 }{ 2^4 }=2^{-4}\]\[64=2^6\]So:\[2^{-4}=(2^6)^{4x-3}=2^{24x-18}\]Now, because these are both equal powers of 2, the exponents should also be equal:

-4=24x-18

Solve it to get x...

Answer 11

Yes, even I understand it now!

Answer 12

x=(14/24)?

Answer 13

Well done, again!

Answer 14

BTW, you may leave out the parentheses now, no need to overdo it :P

Answer 15

Lol I'm parenthesis excited xD
Okay so I know the rules for log problems but how do you solve log equations?

Answer 16

I have to solve
1. 3 log 2x = 4
2. ln(5x + 7) = 8.
3. 5e^(2x=11)=30
And then I'll be done with this one.

Answer 17

Always use:\[\log_{a}x=b \Leftrightarrow a^b=x \]This is the definition of log, in fact.
With it , you can switch back and forth between logarithms and powers.

1.
3 * log(2x)=4, so log(2x)=4/3.
Now remember log is log base 10, so according to the definition, log(2x)=4/3 can also be written as:\[10^{\frac{ 4 }{ 3 }}=x\]
Now there is only some simplification of this last number to do (hint: it is (10^4)^(1/3)

Answer 18

10^4 is 10000

Answer 19

Can you solve it? Is it 21.54?

Answer 20

@Hero can you check if 1 is 21.54?

Answer 21

It is, but don't you have to give an exact answer?

Answer 22

No. But the exact answer is 21.5443469. Can you help with the last two @ZeHanz :)

Answer 23

I meant, \[10^{4/3}=\sqrt[3]{10^4}=\sqrt[3]{10^3 \cdot 10}=10\sqrt[3]{10}\]as an exact answer.

I'll give the other two a try.

Answer 24

I'm not in that advanced yet..

Answer 25

Ok, sou you can round off.

2:
\[\ln(5x+7)=8 \Leftrightarrow e^{8}=5x+7\]
Solve for x and type into your calculator.

Answer 26

Ok, sou you can round off.

2:
\[\ln(5x+7)=8 \Leftrightarrow e^{8}=5x+7\]
Solve for x and type into your calculator.

Answer 27

How do I solve for x with those letters and such? I haven't studied any of this. :/

Answer 28

Shouldn't "2x=11" read "2x-11"?
Then:\[5e^{2x-11}=30\]So\[e^{2x-11}=6\]Uh oh, now the exponent is unknown...
We can solve it (again) by looking at the rule \[\log_{a}x=b \Leftrightarrow a^b=x \]only we use it from right to left instead of the other way round:\[2x-11=\ln 6\]and away you go...

Answer 29

My bad, it's 2x+11 **

Answer 30

Is it 2x+11=ln6 then?

Answer 31

Yes it is!

Sorry, I was getting all excited about problem no. 3.

If you can work with ordinary logs, such as \[\log_{3} x\]or \[\log_{10}x=\log x \]then you need to know that ln is the log with base 2.71828.... = e.

Answer 32

So do you input what e is into the equation to solve?

Answer 33

Or is the solution e^8=5x+7?

Answer 34

Wait, x=(1/5) ln8 @ZeHanz ?

Answer 35

Just use your calculator to get e^8, subtract 7 and divide by 5.