Solving 8log27 and log3x2 = log481/2 without a Calculator

[musicfairy]

Here's 2 I couldn't get.

8^log₂7

log₃x² = log₄8^1/2
Solve for x

For the first one I thought I should solve what's in the exponent first and then solve 8 to that power, but I can't figure out what to do with the exponent.

I'm even more clueless on the second one/ The bases are different, and I can't out what to do with the x² and 8^1/2

Can someone please explain this to me? What properties should I use to solve these problems?

I'm supposed to do these without a calculator, and that's what I'm trying to do.

 

[tiny-tim]

Hi musicfairy!

Romeo and Juliet, Act IV, Scene 4, line 19:

I have a head, sir, that will find out logs​

8^log₂7

Hint: 8 = 2³, and (a^b)^c = … ?

log₃x² = log₄8^1/2
Solve for x

Hint: log₄8 = … ?

 

[musicfairy]

Hi.

So I tried to follow the hints you gave me and came up with these. I might have made a few new properties..

8^log₂7 = 2^3log₂7
So 2 and log₂ kill each other and kill each other and I end up with 21.

For the second one I got

log₄8^1/2 = log₄(2/3)

log₄(3/2) / log₄ 1 = 3/2

2log₃x = 3/2

lnx/ln3 = 3/4

x/3 = e^3/4
x = 3e^3/4

This problem is tricky...

 

[tiny-tim]

Hi musicfairy!

8^log₂7 = 2^3log₂7
So 2 and log₂ kill each other and kill each other and I end up with 21.

Goodness … you musicfairies are violent!

"kill each other"?

Hint: 2^3log₂7 = (2^log₂7)³ = … ?

For the second one I got

log₄8^1/2 = log₄(2/3)

log₄(3/2) / log₄ 1 = 3/2

2log₃x = 3/2

lnx/ln3 = 3/4

x/3 = e^3/4
x = 3e^3/4

Not following any of that

For example, log₄1 = 0, isn't it?

Answer my original question: log₄8 = … ?

 

[musicfairy]

Well, cancel is a better word. I was only quoting someone from class.

2^3log₂7 would equal 7³?

log₄8= 3/2

 

[tiny-tim]

good fairy!

2^3log₂7 would equal 7³?

log₄8= 3/2

Woohoo! ​

And so log₄8^1/2 = … ?

 

[HallsofIvy]

Hi.

So I tried to follow the hints you gave me and came up with these. I might have made a few new properties..

8^log₂7 = 2^3log₂7So 2 and log₂ kill each other and kill each other and I end up with 21.

No, you are correct that the exponential and logarithm with base 2, because they are inverse functions, "kill each other" but you are handling the exponent incorrectly.
8^{log_2(7)}= (2^3)^{log_2(7)}= 2^{3log_2(7)}
But that is NOT 3*7. 3 log₂(7)= log₂(7³).
2^{3 log_2(7)}= 2^{log_2(7^3)}
Now, what is that?

For the second one I got

log₄8^1/2 = log₄(2/3)

Since 8= 2³ and 2= 4^1/2, it is true that 8= 4^3/2 but htat means 8^1/2= 4^(3/2)(1/2)= 4^3/4. And now, log₄(8^1/2= log₄(4^3/4)= 3/4.

log₄(3/2) / log₄ 1 = 3/2

? log₄ 1= 0. log to any base of 1= 0 because a⁰= 1 for any positive number a- and you can't divide by 0. It is certainly NOT true that log(a)/log(b)= a/b !

2log₃x = 3/2

lnx/ln3 = 3/4

x/3 = e^3/4x = 3e^3/4

This problem is tricky...

Is this a different problem now? Are you thinking that e^{ln a/ln b}= a/b? That is definitely not true! It is far easier to use the basic definition of logarithm: If y= log_a(x), then x= a^y. If log₃(x)= 3/4, then x= what?

 

[musicfairy]

Trying to follow your instructions...

log₃x² = 3/4
x² = 3^3/4
x = 3^3/8

Is it right now?

 

[tiny-tim]

x = 3^3/8

Is it right now?

Your fairy logfather says … yes!

 

[musicfairy]

Thanks for all the help and words of wisdom. Now I'll go review log properties.