Question

if 2^2n-1=8^3-n, then the value of n​

Answer 1

$Given \:2^{2n-1} = 8^{3-n}$

$\implies 2^{2n-1} = (2^{3})^{3-n}$

$\implies 2^{2n-1} = (2)^{3(3-n)}$

$\implies 2^{2n-1} = (2)^{9-3n)}$

$\implies 2n - 1 = 9 - 3n$

/* By Exponential Law */

$\boxed { \pink { If \: a^m = a^n\implies m = n }}$

$\implies 2n + 3n = 9 + 1$

$\implies 5n = 10$

$\implies n =\frac{ 10}{2}$

$\implies n = 2$

Therefore.,

$\red{Value \:of \:n } \green {= 2}$

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Answer 2

Answer:

therfore n=2 is the answer

Step-by-step explanation:

2^2n-1=8^3-n

2^2n-1=2^3(3-n). (2^3=8)

bases are equal powers are also equal

2n-1=3(3-n)

2n-1=9-3n

5n=10

n=10/5

n=2