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Solve 5^(3x + 1) = 25^(x + 1) a. x = 4 b. x = -2 c. x = 1 d. x = 0 and could you explain how you got it, so I can understand.
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x = 1
\[5^{3x+1}=25^{x+1}\] = \[5^{3x+1}=5^{2(x+1)}\] =\[equating\ powers\] \[3x+1=2x+2\] \[x=1\]
this is the whole procedure
\[5^(3x+1) = 25^(x+1) = 5^{2\times(x+1)} \] as common base on both sides so,\[3x + 1 = 2(x+1)\] solve for x
\[5^{3x+1}=25^{x+1}, 5^{3x+1}=5^{2(x+1)}, 5^{3x+1}=5^{2x+2}\] \[3x+1=2x+2\] You can substitute and bring 2x on the other side and take 1 on the other side, which leads to this (Also, when you substitute, positive values turn negative, e.g 2x becomes -2x when brought on other side): \[3x-2x=-1+2\] \[\therefore x=1\]
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