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

Robert Marik marik at mendelu.cz
Tue Apr 7 02:13:21 CDT 2009

```黃 裕雄 <bear.tw <at> hotmail.com> writes:

>
>
> /* 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=0as above I use the source
equatioin ,but that can not to solve and
factor.=======================================(%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)=0I let the source eq. => 3^(2*x)-3^x-6=0that
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?==============================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*3^5Can I let that show as follow (2*3)^x*
(2^3)^y=2^8*3^52^(x+3*y)*3^x=2^8*3^5x+3y=8,x=5聰明搜尋和瀏覽網路的免費工具列 —
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>

(and perhaps solvevexplicit)

2. I think that 6^x*8^y=2^8*3^5 has infinitely many solutions. Are x any y reals
or integers?

Robert M.

```