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solve 8 = 2^(x+4)
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2^3=2^(x+4) x+4=3 x=-1
just notice \[\huge 2^{x+4}=2^3 \\ since \\ \huge 8=2^3\]
we can write 8 as 2^3 when bases are equal we can equate powers 2^3=2^(x+4) x+4=3 x=-1
\[8 = 2^{x+4}\] change both sides into same bases first. since \(8 = 2^3\) \[2^3 = 2^{x+4}\] now use the principle \(\text{if} \; \; a^b = a^c \; \text{then} \; b = c\) \[\text{since} \; \; 2^3 = 2^{x+4} \; \; \text{then} \; \; 3 = x+4\] now we solve for x \[3 = x+ 4\] \[3 - 4 = x + 4- 4\] \[3 - 4 = x\] \[-1 = x\]
First try to make the bases same both in L.H.S. & R.H.S as\[2^3=2^{x+4}\]Now if the bases are equal then the powers are also equal so\[3=x+4\]\[-1=x\]
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|dw:1342623693934:dw|thanks! im stuck on this problem too...
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