Use logs to solve 3e7x + 2 = 6 symbolically
- Thread starter awpd229
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If the equation is as you have posted it, then logs are not necessary, since there are no exponents. Did you actually mean any of the following?
. . . . .3<sup>e</sup>7<sup>x</sup> + 2 = 6
. . . . .3e<sup>7</sup>x + 2 = 6
. . . . .3e<sup>7x</sup> + 2 = 6
. . . . .3e<sup>7x + 2</sup> = 6
Or did you mean something else?
When you reply, please show all of the steps you have tried so far. Thank you.
Eliz.
. . . . .3<sup>e</sup>7<sup>x</sup> + 2 = 6
. . . . .3e<sup>7</sup>x + 2 = 6
. . . . .3e<sup>7x</sup> + 2 = 6
. . . . .3e<sup>7x + 2</sup> = 6
Or did you mean something else?
When you reply, please show all of the steps you have tried so far. Thank you.
Eliz.