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JamesJBest ResponseYou've already chosen the best response.1
Well, clearly x = 1 is a solution, yes?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Did you mean: 5^x=4^(x+1)
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
The you could also write :\[5^x=4^{x+1} \Leftrightarrow 5^x=4^x \cdot 4^1 \]
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
yes to \[ 5^x = 4^{x+1} \] ?
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
Now you can divide by 4^x
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
And use \[\frac{ a^x }{ b^x }=\left( \frac{ a }{ b } \right)^x\]
 one year ago

magepker728Best ResponseYou've already chosen the best response.0
dw:1359924788627:dw
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
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.
 one year ago

magepker728Best ResponseYou've already chosen the best response.0
i got to this part
 one year ago

magepker728Best ResponseYou've already chosen the best response.0
@JamesJ that is where my problem is solvinf for that x :D
 one year ago

JamesJBest ResponseYou've already chosen the best response.1
So subtracting ln 4 . x from both sides, we have \[ (\ln 5  \ln 4)x = \ln 4 \] Yes? Now take it from there ...
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
I did the following: \[5^x=4^x \cdot 4 \Leftrightarrow \left( \frac{ 5 }{ 4 } \right)^x=4\]\[x=\log_{\frac{ 5 }{ 4 }} 4=\frac{ \ln4 }{ \ln \frac{ 5 }{ 4 } }=\frac{ \ln 4 }{ \ln5\ln4 }\]
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.1
@JamesJ: sorry, didn't mean to give it away...
 one year ago

magepker728Best ResponseYou've already chosen the best response.0
thank you every1 for ur help
 one year ago
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