There are TWO solutions for x, not just one. To do this you
need to know the change of base formula
and the definition of a logarithm: where c is the exponent to
which base b must be raised to give a. That is is equivalent
to the equation
log3x + logx9 = 3
Use the change of base formula on the second term:
log3x + = 3
Multiply through by log3x
(log3x)² + log39 = 3·log3x
log39 = 2 since the exponent to which the base 3 must be
raised to give 9 is the exponent 2, since 3² = 9
(log3x)² + 2 = 3·log3x
(log3x)² - 3·log3x + 2 = 0
That factors as a quadratic in log3x
(log3x - 2)(log3x - 1) = 0
Use the zero-factor property:
log3x - 2 = 0; log3x - 1 = 0;
log3x = 2; log3x = 1;
x = 32 x = 31
x = 9 x = 3
Two solutions: 9 and 3.
Checking x = 9
log3x + logx9 = 3
log39 + log99 = 3
2 + 1 = 3
3 = 3
Checking x = 3
log3x + logx9 = 3
log33 + log39 = 3
1 + 2 = 3
3 = 3
Both answers check.
Edwin
