Solving 3^x + 3^(x+1) = 4^x

[scientifico]

Homework Statement

3^x + 3^x+1 = 4^x

Can I do this ?

Log 3^x + Log 3^x+1 = Log4^x

thanks

 

[micromass]

No, you can't. The law

$$\log(a+b)=\log(a)+\log(b)$$

is not valid.

 

[Dick]

Homework Statement

3^x + 3^x+1 = 4^x

Can I do this ?

Log 3^x + Log 3^x+1 = Log4^x

thanks

Absolutely not. log(3^x)+log(3^(x+1))=log(3^x*3^(x+1)), not log(3^x+3^(x+1)).

 

[scientifico]

How can I solve my equation ?

 

[tiny-tim]

hi scientifico!

Can I do this ?

Log 3^x + Log 3^x+1 = Log4^x

no!

log(a + b) is not loga + logb !​

hint: factorise 3^x + 3^x+1

 

[Dick]

How can I solve my equation ?

3^(x+1)=3*3^x. Try using that.

 

[scientifico]

Thanks I solved.

 

[scientifico]

3(1-3^x) < 5^x(1-3^x)

must I impose 1-3^x > 0 and 1-3^x < 0 ?

 

[micromass]

3(1-3^x) < 5^x(1-3^x)

must I impose 1-3^x > 0 and 1-3^x < 0 ?

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