Simplifying Logarithms: log_x 32 = 5

In summary, the conversation discusses simplifying an equation involving logarithms and solving it by understanding the definition of a logarithm and using the power rule. The conversation also touches upon using trial and error and the importance of understanding concepts rather than just solving problems.
  • #1
cscott
782
1
[tex]log_x 4 + log_x 8 =5[/tex]

I simplified that to [itex]log_x 32 = 5[/itex] but I can't get my head around what to do next and it's annoying me because I feel it's going to be something simple. :smile:
 
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  • #2
cscott said:
[tex]log_x 4 + log_x 8 =5[/tex]

I simplified that to [itex]log_x 32 = 5[/itex] but I can't get my head around what to do next and it's annoying me because I feel it's going to be something simple. :smile:

What you have isn't that much of a simplification, it would be better to break everything on the LHS to powers of 2 and get smaller numbers. Nevertheless, you can solve the equation immediately from what you have.

What is the definition of a logarithm ? What does it mean when you say [tex]log_a b = c[/tex]. Come up with a simple equation to relate a, b and c using the definition of log, then apply the same principle to your equation, and see what you get.
 
  • #3
hey there since the log is to the base x, it follows that
x^(LOGx32)=x^5 the x and the LOGx cancel out the we have
=>32=x^5
therefore x=2
 
  • #4
Thanks, I understand that now, but I'm having trouble with another :mad:

[tex]20 000 = 10 000(x)^9[/tex]
[tex]2 = x^9[/tex]
[tex]log 2 = log x^9[/tex]
[tex]log 2 = 9 \cdot log x[/tex]

I think I got closer to the answer...
 
  • #5
steven187 said:
hey there since the log is to the base x, it follows that
x^(LOGx32)=x^5 the x and the LOGx cancel out the we have
=>32=x^5
therefore x=2

Is there any way to do it without using trial an error in the end (if the numbers weren't as nice as they are here)?
 
  • #6
cscott said:
Thanks, I understand that now, but I'm having trouble with another :mad:

[tex]20 000 = 10 000(x)^9[/tex]
[tex]2 = x^9[/tex]
[tex]log 2 = log x^9[/tex]
[tex]log 2 = 9 \cdot log x[/tex]

I think I got closer to the answer...
Actually, once you've got [tex]2 = x^9[/tex], all you've got to do is [tex]\sqrt[9]{2} = x = 1.080059739[/tex].

cscott said:
Is there any way to do it without using trial an error in the end (if the numbers weren't as nice as they are here)?
I don't think it's really trial and error; had it been 31 or 33, the answer wouldn't have been an integer. "Luckily", [tex]\sqrt[5]{32} = 2[/tex].
 
  • #7
They design textbook problems so you don't have to go through that mess, just understand the cocnept and know how to repeat the steps you just went through and you'll be fine.
 
  • #8
Just remember, that

[tex]\log_a b[/tex]

is the power which you must raise a to in order to get b. So

[tex]\log_2 32 = 5[/tex]

Equivalently, if

[tex]\log_a b = x[/tex]

then

[tex]b = a^x[/tex]
 

Related to Simplifying Logarithms: log_x 32 = 5

1. What does "log_x 32 = 5" mean?

The notation "log_x 32 = 5" means that the logarithm of 32 to the base x is equal to 5. In other words, x raised to the power of 5 is equal to 32.

2. How do you solve for x in "log_x 32 = 5"?

To solve for x in "log_x 32 = 5", we can rewrite the equation as x^5 = 32 and then take the fifth root of both sides. The solution for x would be the number that, when raised to the fifth power, equals 32.

3. What is the base of the logarithm in "log_x 32 = 5"?

The base of the logarithm in "log_x 32 = 5" is x. This means that x is the number that we are raising to a power in order to get 32.

4. Can the base of the logarithm be any number?

Yes, the base of the logarithm can be any positive number except 1. However, the most commonly used bases are 10 and e (the natural logarithm).

5. What is the value of x in "log_x 1 = 5"?

The value of x in "log_x 1 = 5" is undefined. This is because any number raised to the power of 0 is equal to 1, so there is no solution for x that would make this equation true.

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