If 4log[base]b x - 4log[base]b y = log[base]b z + log[base]b x, express z in terms of x and y.?

Question

I was wondering if anybody could give me a hand in expressing this equation without involving logarithms.

Answer 1

4 log[base b](x) - 4 log[base b](y) = log[base b](z) + log[base b](x)

Your first step is to move constants in front of the log as a power within the log. By this log property,

log[base b](x^4) - log[base b](y^4) = log[base b](z) + log[base b](x)

Next, combine logs on both sides.
On the left hand side, we have a difference of logs, and the difference of logs is the log of their quotient.
On the right hand side, we have a sum of logs, and the sum of logs is the log of their product.

log[base b](x^4 / y^4) = log[base b](zx)

Now, we can take the antilog of both sides and equate the log's insides.

x^4 / y^4 = zx

To express z in terms of x and y, we isolate z by dividing x both sides. This is the same as multiplying (1/x), so I'll do that instead.

(x^4 / y^4) (1/x) = z

And, the 1/x will cancel with one x on the top, making the final answr

z = x^3 / y^4

Answer 2

Rewrite equation:
log b (x^4) - log b (y^4) = log b z+ log b x
use these laws:
log a +log b= log(a*b)
log a -log b= log(a/b)
where base is n ( n>1 and belongs to Z)

log b (x^4/y^4)= log b (z*x)
take anti log of both sides
b^(x^4/y^4)=b^(z*x)
same base therefore
x^4/y^4=z*x
x^3=z*y^4

Answer 3

do no longer ignore that log(a) + log(b) = log(ab) log 324 = log(2*2*3*3*3*3) -- the bottom b is redundant log 324 = log 2 + log 2 + log 3 + log 3 + log 3 + log 3 log 324 = 2log 2 + 4log 3 log 324 = 2x + 4y do no longer ignore that log(a) - log(b) = log(a/b) log 8/27 = log(2*2*2/3/3/3) log 8/27 = 3log 2 - 3log 3 log 8/27 = 3x - 3y log 9 = 2log 3, log 8 = 3log 2 log 9 / log 8 = 2y / 3x

Answer 4

1. 4log[base]b x - 4log[base]b y = log[base]b z + log[base]b x
2. 4log (base)b (xdivdey) = log (base)b zx
3. (xdividey) to the power of 4 = zx
4. z = (x to the power of 3 /
divide y to the power of 4)
QED

Answer 5

Let's assume all logs to be base b

4logx-4logy-logx=logz

3logx-4logy=logz

log(x^3) - log(y^4) = logz

log(x^3/y^4) = log z

z = x^3/y^4

Answer 6

If 4log[base]b x - 4log[base]b y = log[base]b z + log[base]b x

=>4(log[base]b (x/y))=log[base]b (xz)

since log (ab)=log a + log b
and log(a/b)=log a - log b

=>log[base]b ((x/y)^4)=log[base]b (xz)

Since base is same on both sides
Now, above becomes
(x/y)^4=xz
=>z=(x^3)(y^4)