Question

${6}^{15} \times {10}^{5} = {2}^{x} \times {3}^{y} \times {5}^{z} find(x + y + z)$
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Answer 1

Step-by-step explanation:

${6}^{15} \times {10}^{5} = {2}^{x} \times {3}^{y} \times {5}^{z} \\ \\ = > {2}^{15} \times {3}^{15} \times {5}^{5} \times {2}^{5} = {2}^{x} \times {3}^{y} \times {5}^{z} \\ \\ = > {2}^{20} \times {3}^{15} \times {5}^{5} = {2}^{x} \times {3}^{y} \times {5}^{z} \\ \\ compare \: = > \\ \\ \\ \\ x = 20 \: \: \: \: \: y = 15 \: \: \: \: \: z = 5 \\ \\ x + y + z = 20 + 15 + 5 = 40$

Answer 2

Answer:

$\ \: we \: have \\ \ \: \\ \implies \: \red{ {6}^{15} \times {10}^{5} = {2}^{x} \: \times {3}^{y} \times {5}^{z} }\\ \\ \implies \: {(2 \times 3)}^{15} \times {(2 \times 5)}^{5} = {2}^{x} \times {3}^{y} \times {5}^{z} \\ \\ \implies \: {2}^{15} \times {3}^{15} \times {2}^{5} \times {5}^{5} = {2}^{x} \times {3}^{y} \times {5}^{z} \\ \\ \implies \: \green{ {2}^{20} \times {3}^{15} \times {5}^{5} = {2}^{x} \times {3}^{y} \times {5}^{z} }\\ \\ \red{ \bf{ compairing \: on \: both \: sides}} \\ \\ \therefore \\ \: \red{\boxed{ x = 20}} \: , \: \: \green{\boxed{y = 15}} \: , \: \pink{\boxed{z = 5}}$

Now,

=X+y+z

= 20+15+5

= 40 $\mathfrak{Answer}$