Log Equations

[mathxyz]

How do I solve equations like this:

Solve for x: log_16x + log_4x + log_2x = 7

Notice: Every log has a DIFFERENT subbase: (16, 4 and 2) See it?

 

[tkhunny]

Use the property below to transform all the logs to the same base.

log_a(b) = log_c(b)/log_c(a)

My first guess would be to try c = 2.

 

[mathxyz]

tkhunny

I never saw this conversion formula for logs before. Can you do the first one for me and than I will follow from there on?

 

[soroban]

Hello, mathxyz!

Solve for x: .log₁₆(x) + log₄(x) + log₂(x) .= .7
.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . log(N)
Base-change Formula . .log_b(N) . = . --------- . . (use any base here)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . log(b)


. . . . . . . . . . . . . . . . . . . . .log(x) . . . . . log(x) . . . . . .log(x)
We have: . log₁₆(x) . = . ---------- . = . --------- . = . ----------
. . . . . . . . . . . . . . . . . . . . log(16) . . . . log(2⁴) . . . . 4 log(2)

. . . . . . . . . . . . . . . . . . . log(x) . . . . .log(x) . . . . . log(x)
. . . . and: . log₄(x) . = . -------- . = . --------- . = . ----------
. . . . . . . . . . . . . . . . . . . log(4) . . . . log(2²) . . . .2 log(2)

. . . . . . . . . . . . . . . . . . . log(x)
. . . . and: . log₂(x) . = . --------
. . . . . . . . . . . . . . . . . . . log(2)


. . . . . . . . . . . . . . . . . . . . . log(x) . . . .log(x) . . . log(x)
The equation becomes: . ---------- + ---------- + --------- . = . 7
. . . . . . . . . . . . . . . . . . . . 4 log(2) . . 2 log(2) . . log(2)

. . . . . . . . . . . . . .7 log(x)
Then we have: . ----------- . = . 7
. . . . . . . . . . . . . .4 log(2)

Hence: . log(x) . = . 4 log(2) . = . log(2⁴) . = . log(16)

Therefore: . log(x) .= .log(16) . . ---> . . x .= .16

 

[mathxyz]

Hey..

Thank you for for your help.

TO Soroban:

Good to hear from you again. How was your trip?

 

[tkhunny]

Re: tkhunny

I never saw this conversion formula for logs before.

Yikes! I wonder how you were expected to solve such problems? Definitely something wrong with the structure of the book or curriculum on that one. I'm glad you asked, so we could clear it up.

 

[mathxyz]

yes

There is somehting very worng with the USA educational system.

 

[tkhunny]

Oh, I don't think we can blame your experience on the whole of the United States of America. I think you have a rather specific and esoteric example. Really, I would love to see the curriculum and read the goals of the sponsoring institution.

 

[soroban]

Hello, mathxyz!

There is a way to solve this without the Base Change Formula, but it's also pretty messy.
. . If you can't follow it, don't lose any sleep over it, okay?

Solve for x: .log₁₆x + log₄x + log₂x = 7
.

Consider the middle term: . log₄x
. . Let log₄x = a . ---> . 4^a = x

Consider the first term: . log₁₆x = b . ---> . 16^b = x
. . We have: . (4²)^b = x . ---> . 4^2b = x . ---> . log₄x .= .2b .= .2·log₁₆x
We've just shown that the middle term is twice the first term.


Consider the last term: . log₂x = c . ---> . 2^c = x

Consider the middle term: . log₄x = a . ---> . 4^a = x
. . We have: . (2^2)^a = x . ---> . 2^2a = x . ---> . log₂x .= .2a .= .2·log₄x
We've just shown that the last term is twice the middle term, or four times the first term.

The equation becomes: . log₁₆x + 2·log₁₆x + 4·log₁₆x . = . 7

And we have: . 7·log₁₆x .= .7

. . . . . . Then: . . . log₁₆x .= .1

. . .Therefore: . . . . . . . x .= .16¹ .= .16