## magepker728 Group Title 1help step by step please on how to do this 5^x=4^x+1 one year ago one year ago

1. JamesJ

Well, clearly x = 1 is a solution, yes?

2. ZeHanz

Did you mean: 5^x=4^(x+1)

3. magepker728

yes

4. ZeHanz

The you could also write :$5^x=4^{x+1} \Leftrightarrow 5^x=4^x \cdot 4^1$

5. JamesJ

yes to $5^x = 4^{x+1}$ ?

6. magepker728

yes @JamesJ

7. ZeHanz

Now you can divide by 4^x

8. ZeHanz

And use $\frac{ a^x }{ b^x }=\left( \frac{ a }{ b } \right)^x$

9. magepker728

|dw:1359924788627:dw|

10. JamesJ

Alternatively, take the natural log of both sides and you have $\ln 5^x = \ln 4^{x+1}$ $x \ln 5 = (x+1) \ln 4$ Now solve for x.

11. magepker728

i got to this part

12. magepker728

@JamesJ that is where my problem is solvinf for that x :D

13. JamesJ

So subtracting ln 4 . x from both sides, we have $(\ln 5 - \ln 4)x = \ln 4$ Yes? Now take it from there ...

14. ZeHanz

I did the following: $5^x=4^x \cdot 4 \Leftrightarrow \left( \frac{ 5 }{ 4 } \right)^x=4$$x=\log_{\frac{ 5 }{ 4 }} 4=\frac{ \ln4 }{ \ln \frac{ 5 }{ 4 } }=\frac{ \ln 4 }{ \ln5-\ln4 }$

15. ZeHanz

@JamesJ: sorry, didn't mean to give it away...

16. magepker728

thank you every1 for ur help