Question

... sometime somewhere I found this ...

any idea how can i solve it ?

Answer 1

x = log_a bc
y = log_b ca
z = log_c ab

1 + abc = ?

Answer 2

I am sorry it's beyond my capability to solve such a difficult problem.

Answer 3

Well for a basic idea, we just have to isolate A,B,and C somehow from X,Y, and Z and multiply them together and add 1

Answer 4

hope somebody math specialist will can help me with any ideas

Answer 5

Is this for an actual assignment?

Answer 6

Well first for the first one (x=), we can isolate A by dividing BC by both sides making the equation X/BC=log_a. Now what cancels out log? I'm kinda sure this is how you do it.

Answer 7

i ve got using the rules of logarithms that

x = log_a bc => bc = a^x
y = log_b ca => ca = b^y
z = log_c ab => ab = c^z

how continue it ?

Answer 8

no @ramen in case of logarithms this more complicated bc here log_a so a mean the index the base of logarithm

Answer 9

Yea that's kinda what I thought what was wrong

Answer 10

@Hero

Answer 11

I'm not 100& sure rn but I will try to figure it out.

Answer 12

*%

Answer 13

bc = a^x
ca = b^y
ab = c^z

so than we multiplie both sides term by term we get

bc*ca*ab = a^x *b^y *c^z

a^2 *b^2 *c^2 = a^x *b^y *c^z

and now ?

Answer 14

@dude

Answer 15

@imqwerty

Answer 16

hmm... are we trying to solve this just in terms of x, y and z?

Answer 17

Here is what I tried doing

x = log_a bc
y = log_b ca
z = log_c ab

So then using change of base formula for logs

x = log(bc)/log(a)
y = log(ca)/log(b)
z = log(ab)/log(c)

then isolating for a, b and c in the equations

xlog(a) = log(bc)
ylog(b)=log(ca)
zlog(c) = log(ab)

so then log(a) = log(bc)/x
log(b) = log(ca)/y
and log(c) = log(ab)/z

When log is written in that form we assume base 10, so to cancel out the log we can raise it to the exponent of 10

So a = 10^(log(bc)/x)
b = 10^(log(ca)/y)
c = 10^(log(ab)/z)

Then maybe multiply those 3 together and add 1.
But prolly want to take a different approach is we are only using x, y and z as the only variable

Let me try that now and see if I can get a lead

Answer 18

Yeah, probably disregard that last step if you want the answer in terms of x, y and z.
So we know what a b and c equal too. From there I would assume the next step is to use a system of equations. 3 variables and 3 equations. We know what a b and c equal to so then you would substitute the value of 1 variable into the other equations and also use elimination wherever possible to get the answer in terms of a, b and c. Though I do not know how this would end up or even if it is possible.

Answer 19

Also you can google up your problem and you will get some similar problems, try to get an idea from that as well

Answer 20

@imqwerty

Answer 21

we want to find the value for 1 + abc in terms of x,y,z i guess?

one simple (but ugly) way to do this-

convert the equations so they look like this-

\(x = log_a(bc) \)

\(x = \large\frac{log(bc)}{log(a)}\)

\(x log(a) = log(bc)\)
\(x log(a) = log(b) + log(c)\)

I'm renaming \(log(a) ~ log(b) ~ log(c)\) as A, B, and C respectively.
If we have A + B + C, then we can also get 1 + abc from it.
so our new task is to find A + B + C.

the three equations convert to:
\(xA = B + C\)
\(yB = C + A\)
\(zC = A + B\)

three equations, three unknowns. System of linear equations