Question

If [(9/7)^5x-3]^2=6561/2401.find x​

Answer 1

Tip:

Recall the properties of exponents

• If $a^m=a^n$ then $m=n$
• $(a^m)^n=a^{mn}$

Given: $[(\frac{9}{7} )^{5x-3}]^2=\frac{6561}{2401}$

To find: The value of $x$

Using the property $(a^m)^n=a^{mn}$,

$\implies (\frac{9}{7} )^{10x-6}=\frac{6561}{2401}$

$6561$ can be written in exponential form as $9^4$

$2401$ can be written in exponential form as $7^4$

$\implies (\frac{9}{7} )^{10x-6}=\frac{9^4}{7^4}$

$\implies (\frac{9}{7} )^{10x-6}=(\frac{9}{7})^4$

Since the base is the same, equate the exponents

$\implies 10x-6=4$

$\implies 10x=6+4$

$\implies 10x=10$

$\implies x=1$

The value of $x=1$

Answer 2

Given: $[(\frac{9}{7})^{5x-3}]^2=\frac{6561}{2401}$

Taking Square roots on both sides,

$(\frac{9}{7})^{5x-3}=\sqrt{\frac{6561}{2401}}$

$(\frac{9}{7})^{5x-3}=\frac{81}{49}$

$(\frac{9}{7})^{5x-3}=[\frac{9}{7}]^2$

The base of the power is the same on both sides. Hence we can equate the power.

$5x-3=2\\5x=5\\x=1$

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