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[c-duced]

Can someone please solve this for me pleasE? -___-'

ln(x+2)/ln(2)-2^x+2 = 0

 

[Slidey]

ln(x+2)-ln(4^x)+ln(4)=0
ln(x+2)=ln(4^x/4)
x+2=4^(x-1)
Hmm. By inspection, x=2, but...
OK: ln(x+2)-ln(4^x/4)=0
ln([x+2]/[4^(x-1)])=0
(x+2)(4^[x-1])=1
Uhm.
Blow it. Just be happy with x=2, although I know that's not the only solution.

 

[c-duced]

x=-1?.......

btw, thanks neway

 

[Xayma]

No approx -1.69

ln (x+2)/ln (2)-2^x+2=0
log₂ (x+2)-2^x+2=0
log₂ (4x+8)=2^x
log₂ (4x+8)=2^x
Thats where I got to.

 

[Slidey]

Indeed. There could be more, but I would think x=-1.69 and 2 would suffice.

 

[Xayma]

Indeed. There could be more, but I would think x=-1.69 and 2 would suffice.

No those are the only two.

If we consider the graphs y=log₂ (4x+8) and y=2^x

They only intersect a maximum of two times. Therefore there is two roots (consider the log graph and exponential graph and convince yourself of this).

 

[Slidey]

Indeed. I wasn't thinking.

 

[Archman]

No those are the only two.

If we consider the graphs y=log₂ (4x+8) and y=2^x

They only intersect a maximum of two times. Therefore there is two roots (consider the log graph and exponential graph and convince yourself of this).

no, the proper way to say this is that one graph is strictly convex while the other is strictly concave, so they can only meet at a maximum of 2 places.

 

[Xayma]

I was trying to say something like that, but couldn't word it. Plus it is irrelevant here, he isn't a HSC marker and can understand each other.

And I couldn't be bothered finding d²y/dx² (too much calculus when surely he should know the standard log and exponential graphs.

 

[~ ReNcH ~]

Can someone please solve this for me pleasE? -___-'

ln(x+2)/ln(2)-2^x+2 = 0

Can you actually solve that equation algebraically? It's one of those "transcendental" equations isn't it?

 

[Slidey]

It looks to be transcendental to me.

I haven't done them yet, but perhaps Newton's root methods would help. Maybe that's how Xayma got the other solution.

 

[Xayma]

Stuff that, I didnt want to do the derivative. I manually just found it to 2 dp it was quick.