# [Maxima] how to use the some base to solve exponential fuction

=?big5?B?tsAguM62rw==?= bear.tw at hotmail.com
Tue Apr 7 00:31:28 CDT 2009

```
/* use the "same base" in exponential function*/

solve the source eq. 9^x-3^x-6=0
(%i35) 9^x-3^x-6=0;
(%i36) solve([%], [x]);
(%o36) [9^x=3^x+6]

(%i37) 9^x-3^x-6=0;
(%i38) factor(%);
(%o38) 9^x-3^x-6=0

as above I use the source equatioin ,
but that can not to solve and factor.
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(%i40) 3^(2*x)-3^x-6=0;
(%i41) solve([%], [x]);
(%o41) [x=1,x=log(-2)/log(3)]

(%i42) 3^(2*x)-3^x-6=0;
(%i43) factor(%);
(%o43) (3^x-3)*(3^x+2)=0

I let the source eq. => 3^(2*x)-3^x-6=0
that can be solve and factor,
but the one of solution "x=log(-2)/log(3)" is not correct represention.
(since log(x) ,where x must be positive number).

If I want to use the source eq. " 9^x-3^x-6=0"
How can I solve and factor directly?

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2.
I want to solve 9^x=27^(x-1)
Can I let that show as follow

(3^2)^x=(3^3)^(x-1)
3^(2*x)=3^(3*(x-1))

3.
I want to solve 6^x*8^y=2^8*3^5
Can I let that show as follow

(2*3)^x*(2^3)^y=2^8*3^5
2^(x+3*y)*3^x=2^8*3^5
x+3y=8,x=5

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