+0  
 
0
338
6
avatar

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+7347 
+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+94548 
+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+94548 
+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+7347 
+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+94548 
+1

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

 

 

cool cool cool

 Nov 5, 2017
 #6
avatar+20848 
+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

24 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.