Trying to workout 3(3$^x$) = 27$^{2x}$
So far i have done the following,
3(3$^x$)= 27$^{2x}$ (3)(3$^x$) = 27$^{2x}$ (3)+$x$= (9$^x$)(3$^x$)
Cant see what to do next...
Thanks for any help!
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you need to use the inverse of the exponential called a logarithm https://en.m.wikipedia.org/wiki/Logarithm – shai horowitz Oct 09 '16 at 13:54
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Appreciaite the reply. I think the poster below solved it without using the log but I can see where your coming from in regards using it. Thanks – John Allen Oct 09 '16 at 14:03
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technically he used the log setting x+1=6x – shai horowitz Oct 09 '16 at 15:40
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$$3\cdot \left( { 3 }^{ x } \right) ={ 27 }^{ 2x }\\ { 3 }^{ x+1 }={ \left( { 3 }^{ 3 } \right) }^{ 2x }\\ { 3 }^{ x+1 }={ 3 }^{ 6x }\\ x+1=6x\\ x=0.2$$
haqnatural
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