How do you solve this sorta equations:

b^x = ax?

Specifically, this example :

(1.09)^x = 0.546x

Question

How do you solve this sorta equations:

b^x = ax?

Specifically, this example :

(1.09)^x = 0.546x

anonymous · Answer

damn it's different !

anonymous · Answer

i think we have to equal the base then when base is equal we can just ignore the bases so we will try to solve the exponents for x
for exmple: 25^x=5^x-1
    we can write 25 like this 5^2 so we will write instate of  25:
5^2x=5^x-1    now because the bases are equal so we will ignore them we will solve the exponent for x 
2x=x-1
x=-1
hope i could help you

anonymous · Answer

@Roh its  b^x = ax and not b^x = a^x

anonymous · Answer

@mridul1 so dear b^x=ax   so we will get that x=1 then we will solve it ..

anonymous · Answer

Look in Examples section here: http://en.wikipedia.org/wiki/Lambert_W_function

anonymous · Answer

hey thanks people. looks complicated indeed. is there any easy way to evaluate the W function? In this case, 
x = -W(-1.67) / ln(2) .. now?

anonymous · Answer

Use the Taylor series expansion on the same page (or Wolfram).

anonymous · Answer

right! thanks dude.. you rock

anonymous · Answer

ur welcome.