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PLLLEEZZZ! solve: 7^(x-5)=7^(4x+7)
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take log of base 7 on both sides to get x-5 = 4x + 7
\[\large x^a = x^b \implies a = b\]
No need to take log here..
Just equate the exponents...
i dont understand...
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since bases are same so just equate both sides \[\Large (x-5)=4x+7\] take all the x terms to one side \[\Large 4x-x=-12\] \[\Large 3x=-12\] i hope you got it.
Tell me what are the exponents both the sides ??
@cassie4math did you get it?
I give you the example: \[\huge 5^{x-6} = 5^{-2x}\] This implies : \[\large x - 6 = -2x\]
so x=-4 riught?
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Well Done..
thanks!
Welcome dear..
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