Using logs find x

5(to the power of x) = 17

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`5^x=17`

Taking log on both side

`log(5^x)=log(17)`

Since `loga^m=mloga`

Therefore

`xlog5=log17`

`x=log(17)/log5`

`x approx 1.7604 `

The equation `5^x = 17` has to be solved for x.

Take the logarithm to base 10 of both the sides

`log_10(5^x) = log_10 17`

Use the relation `log_10 a^b = b*log_10 a`

`x*log_10 5 = log_10 17`

`x = (log_10 17)/(log_10 5)`

The approximate value of x is 1.76

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