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x^a=y^b=k and x^c=y^d=t, then which of the following is true. 

 

a) ac=bd

b) ad=bc

c) a/d=c/b

d) a^c=b^d

e) a+c=b+d

 Nov 5, 2017
 #1
avatar+7612 
+1

x^a  =  y^b

 

(x^a)^n  =  (y^b)^n

 

x^(an)  =  y^(bn)

 

Let's say

 

c  =  an     and     d = bn

 

c/a  =  n     and     d/b  =  n

 

c/a  =  d/b     cross multiply

 

ad  =  bc

 Nov 5, 2017
 #2
avatar+100569 
+2

x^a  = k   → log x  =  log k / a

y^b  = k   → log y  =  log k / b

x^c  = t   

y^d  = t

 

So

 

k * t   =  x^a * x^c =  y^b * y^d  =  x^(a + c) = y^(b + d)

 

So

 

(a + c) log x  =  ( b + d) log y

 

(a + c) ( log k / a)  =  (b + d)( log k / b)

 

(a + c) b  =  ( b + d) a

 

ab + bc  =   ab + ad

 

[ bc   =  ad ]  →  [ ad   =  bc ]

 

 

 

cool cool cool

 Nov 5, 2017
 #3
avatar+100569 
+1

Sorry, hectictar....I didn't see you working on this at the same time !!!

 

Different procedures....same answer  !!!!

 

 

 

cool cool cool

 Nov 5, 2017
 #4
avatar+7612 
+1

I'm glad you answered it too and got the same thing....I was a little bit unsure of my answer!!!

hectictar  Nov 5, 2017
 #5
avatar+100569 
+1

Maybe we're both wrong, but think that we're correct.....LOL!!!!!

 

 

cool cool cool

 Nov 5, 2017
 #6
avatar+22188 
+1

x^a=y^b=k and x^c=y^d=t, then which of the following is true.

 

a) ac=bd

b) ad=bc

c) a/d=c/b

d) a^c=b^d

e) a+c=b+d

 

\(x^a=y^b \\ x^c=y^d\)

 

\(\begin{array}{|rcll|} \hline x^a &=& y^b \quad & | \quad \text{exponentiate both sides with } \frac{c}{a} \\ \displaystyle x^{a\cdot\frac{c}{a}} &=& \displaystyle y^{b\cdot\frac{c}{a}} \\ x^{c} &=& \displaystyle y^{b\cdot\frac{c}{a}}\quad & | \quad x^{c} = y^d \\ y^d &=& \displaystyle y^{b\cdot\frac{c}{a}} \\ \Rightarrow d &=&\displaystyle b\cdot\frac{c}{a} \\ \mathbf{ad} & \mathbf{=} & \mathbf{bc} \\ \hline \end{array} \)

 

laugh

 Nov 7, 2017

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