Expert Answers
durbanville eNotes educator| Certified Educator

To solve without using logs, once you get to the stage of :

`7.2^(x^2) - 2.7 ^(x^2) = 0` 

`therefore 7.2^(x^2) = 2.7^(x^2)`

`therefore (2^(x^2))/(7^(x^2))= 2/7`  having rearranged the equation

`therefore (2/7)^(x^2) = 2/7`

Now that the bases are the same, the powers/ exponents will also equal each other

`therefore x^2=1`

therefore x = 1 or x = -1

llltkl | Student


`=> 7*(2^2)^(x^2)-9*(7*2)^(x^2)+2*(7^2)^(x^2)=0`

`=> 7.2^2^(x^2)-7.(7.2)^(x^2)-2.(7.2)^(x^2)+2.7^(2)^(x^2)=0`

` ` `rArr 7*2^(x^2)(2^(x^2)-7^(x^2))-2*7^(x^2)(2^(x^2)-7^(x^2))=0`

`rArr (2^(x^2)-7^(x^2))(7*2^(x^2)-2*7^(x^2))=0`

if, `(2^(x^2)-7^(x^2))=0` `rArr 2^(x^2)=7^(x^2)`

then x must be equal to 0

if, `(7*2^(x^2)-2*7^(x^2))=0`

then `7*2^(x^2)=2*7^(x^2)`

Taking log on both sides we get,

log7+`x^(2)` log2=log2+`x^(2)` log7

or, `x^(2)` (log2-log7)=log2-log7

or, `x^(2)` =1

or x=`+-1`

Therefore, the values of x are 0,+1,-1.