Question

Solve the following

$\sf\ if \: \ a^x=b\ , \: \ b^y= c\ ,\ \: c^z=d \\ \\\sf\ find \: \: the \: value \: of \: \: xyz \\ \\ \sf\ a) \: 1 \: \: \ \: \: \ \ b) \: 0 \: \ \: \: \ \ c) \: \dfrac{1}{abc}\ \: \ \: \: \: \ d) \: abc$​

Answer 1

Correct Question :

If a^x = b ; b^y = c and c^z = a , find the value of xyz.

Answer :

Option A ; xyz = 1

Solution :

Here, for 7 numbers , a, b , c , d, x, y and z € R the following properties hold .

1. a^x = b

2. b^y = c

3. c^z = a

Method 1 :

Multiply all the values on the LHS and the RHS

=> [ a^x ] [ b^y ] [ c^z ] = abc

=> { abc }^{ xyz} = abc

=> xyz = 1 .

This is the required answer.

Method 2 :

a^x = b

=> log ( a^x) = log b

=> x log a = log b ...... (1)

b^y = c

,=> log ( b^y) = log c

=> y log b = log c ...... (2)

c^z = a

=> log c^z = log a

=> z log c = log a. ...... (3)

From 1 :

=> [ log b ]/[ log a] = x

Similarly

=> [ log c ]/[ log b] = y

=> [log a]/[ log c ] = z

Multiplying all these

The terms on the LHS cancel out leaving us with xyz = 1.

This is the required answer.

________________________________________

Answer 2

${\large{\bold{\sf{\underline{Question}}}}}$

$\sf\ if \: \ a^x=b\ , \: \ b^y= c\ ,\ \: c^z=d \\ \\\sf\ find \: \: the \: value \: of \: \: xyz \\ \\ \sf\ a) \: 1 \: \: \ \: \: \ \ b) \: 0 \: \ \: \: \ \ c) \: \dfrac{1}{abc}\ \: \ \: \: \: \ d) \: abc$

${\large{\bold{\sf{\underline{Given \; that}}}}}$

➨ aˣ = b

➨ bʸ = c

➨ cᶻ = d

${\large{\bold{\sf{\underline{To \; find}}}}}$

➨ The value of ˣʸᶻ

${\large{\bold{\sf{\underline{Solution}}}}}$

➨ The value of ˣʸᶻ = 1

${\large{\bold{\sf{\underline{Full \; Solution}}}}}$

• Let's multiply ( RHS - LHS )

➨ ( aˣ ) ( bʸ ) ( cᶻ ) = abc

➨ abcˣʸᶻ = abc

• Let's cancel abc by abc

➨ ˣʸᶻ = 1

${\large{\bold{\sf{\underline{Additional \; formulas}}}}}$

Algebric Identities :

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$

Law's of exponents :

$\begin{gathered}\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}\end{gathered}$