Question

What will be the value of z satisfying the equation 7^x× 9^y× 4^z = 14^4× 6^8?​

Answer 1

Given :   $7^x\times9^y\times4^z=14^4\times 6^8$

To Find : Value of z

Solution:

$7^x\times9^y\times4^z=14^4\times 6^8$

14 = 2 * 7

6 = 2 * 3

9 = 3 * 3 = 3²

4 = 2 * 2 = 2²

xᵃ * xᵇ = xᵃ⁺ᵇ

(xᵃ)ᵇ = xᵃᵇ

$7^x\times(3^2)^y\times(2^2)^z=(2 \times7)^4\times (2 \times 3)^8$

$7^x\times3^{2y}\times2^{2z}=2^4 \times7^4\times 2^8 \times 3^8$

$7^x\times3^{2y}\times2^{2z}= 7^4 \times 3^8 \times 2^{12}$

as 2 , 3 and 7 are prime numbers

Hence powers for equal bases can be equated

x = 4

2y = 8  => y = 4

2z = 12 => z = 6

Value of z = 6

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