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1help step by step please on how to do this
5^x=4^x+1



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

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

magepker728
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yes

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

JamesJ
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yes to
\[ 5^x = 4^{x+1} \]
?

magepker728
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yes @JamesJ

ZeHanz
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Now you can divide by 4^x

ZeHanz
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And use \[\frac{ a^x }{ b^x }=\left( \frac{ a }{ b } \right)^x\]

magepker728
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dw:1359924788627:dw

JamesJ
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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.

magepker728
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i got to this part

magepker728
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@JamesJ that is where my problem is solvinf for that x :D

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

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

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

magepker728
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thank you every1 for ur help