Question

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7^X-1= 1 SOLVE THIS EQUATION PLEASE

Answer 1

$Given : 7^{x-1} = 1$

Apply log on both sides, we get

$= \ \textgreater \ log( 7^{x-1}) = log(1)$

We know that log(m^n) = n log m

$= \ \textgreater \ (x - 1) log 7 = log 1$

We know that log 1 = 0

$= \ \textgreater \ (x - 1) log 7 = 0$

$= \ \textgreater \ x - 1 = \frac{0}{log 7}$

$= \ \textgreater \ x - 1 = 0$

$= \ \textgreater \ x = 1.$


Therefore the value of x = 1.


Hope this helps!