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Evern Group TitleBest ResponseYou've already chosen the best response.0
\[5^{(x+1)}(13) = 22 =>5^{(x+1)} = 11 =>x is imaginary\]
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
1 certainly works
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[5^23=253=22\]
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
you can write \[5^{x+1}=5^2\times 5^{x1}\] and the solve \[25\times 5^{x1}3\times f^{x1}=22\] \[22\times 5^{x1}=22\] \[5^{x1}=1\] \[x1=0\] \[x=1\]
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
that "f" in line 4 should be a "5"
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
btw 1 is not imaginary. just thought i would mention it
 3 years ago

Evern Group TitleBest ResponseYou've already chosen the best response.0
sorry .. i misread the second exponent as x+1
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
oh, guess there would be nothing then for sure!
 3 years ago

majamin Group TitleBest ResponseYou've already chosen the best response.0
Another version: \[5^{x+1}  3\cdot5^{x1} = 22\] \[5\cdot 5^x\frac{3}{5}5^x=22\] \[25\cdot 5^x3\cdot 5^x=110\] \[22\cdot 5^x=110\] \[5^x=5\] \[x=1\]
 3 years ago
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