- Thread starter
- #1

- Thread starter siyanor
- Start date

- Thread starter
- #1

- Mar 1, 2012

- 249

The first method is not correct because ln(7.3) has simply disappeared.I have an equation which answered it in two ways,which one is correct ?

View attachment 163

The second

- Apr 3, 2012

- 37

[tex]7.3^{2x + 1} \ = \ 10[/tex]I have an equation which answered it in two ways,which one is correct ?View attachment 163

siyanor,

(regarding the correct process of method #2),

because you have 10 on one side at that point,

that lends itself to taking log (base 10) of each

side:

[tex]\log_{10}7.3^{2x + 1} \ = [/tex][tex] \log_{10}10[/tex]

[tex](2x + 1)\log(7.3) \ = \ 1[/tex]

[tex]2x + 1 \ = \ \dfrac{1}{log(7.3)}[/tex]

And continue . . .