# how would i be able to graph f(x)=4log (base 2) x+ 6?any help os greatly appreciated!(:

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Graph `f(x)=4log_2x+6` .

(1) One way is to graph `f(x)=log_2x` , then use transformations to get the graph of `f(x)=4*log_2x+6` .

`f(x)=log_2x` is defined for x>0. It is an ever increasing curve, but the rate of increase slows down as x grows. It goes through the points (1/4,-2),(1/2,-1),(1,0),(2,1),(4,2),(8,3), etc...

Multiplying by 4 has the effect of a vertical stretch; each y-value is multiplied by 4. So the graph of `y=4*log_2x` goes through the points (1/4,-8),(1/2,-4),(1,0),(2,4),(4,8),(8,12),etc...

Finally adding six is a vertical translation up 6 units; add six to each y-value. **Thus the graph of `f(x)=4*log_2x+6` goes through the points (1/4,-2),(1/2,2),(16),(2,10),(4,14),(8,18),etc...**

(2) Another way to graph is to graph `f(x)=(4logx)/(log2)+6` where `log` is the common logarithm, base 10. (Using the change of base formula: `log_cb=(logb)/(logc) "or" (lnb)/(lnc)` )

The graph: