this is solve for x can anyone out there help me please

## 2^x+2+2^x+1=192

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# 2 Answers

What if the problem is really 2^(x+2)+2^(x+1)=192 ?

If this is the case we have (2^x)(2^2)+(2^x)(2^1)=192

(2^x)4+(2^x)2=192

6(2^x)=192

2^x=32

x=5(we don't need logs because we know 32 is a power of 2)

check: (2^7)+(2^6)=128+64=192

This problem is identical in form to the last one that you asked.

you have some:

k

^{x}+k^{x+1}=cTo solve equations of these forms you must recognize that:

k

^{x+1}=k(k^{x})this allows you to re-express the original equation as:

k

^{x}+k(k^{x})=cfactoring you get

k

^{x}(k+1)=ck

^{x}=c/(k+1)x=log

_{k}(c/(k+1))Specifically, here you have k=2, and c=192, so:

x=log

_{2}(192/3)x=log

_{2}(64)x=6