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Solve 7^(x-5)=7(^4x+7)
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A. x=-4 B. x=-12/5 C. x=2/3 D. x=4
is the problem \[ 7^{(x-5)}=7^{(4x+7)}\]
Yes!
I know ^x equals (x+2) but I'm confused about the 7s there
If the bases are equal (and they are both 7) then to make both sides equal you need to make the exponents equal to each other in other words, you need to find x so that \[ x-5= 4x+7 \]
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So i just move the x over and minus and add like normal?
I would first add -x to both sides (and simplify) then add -7 to both sides (and simplify)
x-x−5=4x-x+7
so -12/3?
that simplifies
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-4!!!
thankyou!!!
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