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C2 Logs question

How would I answer 3 x 2^x = 7^x-2. I keep trying and getting -29.145, which is obviously wrong. Thanks in advanceIMG_5947.jpg
Original post by doglover123
How would I answer 3 x 2^x = 7^x-2. I keep trying and getting -29.145, which is obviously wrong. Thanks in advanceIMG_5947.jpg


I'm sorry the photo is sideways, i have no idea how to rotate it :frown:
Bring the x terms to one side and factorise.
Original post by doglover123
How would I answer 3 x 2^x = 7^x-2. I keep trying and getting -29.145, which is obviously wrong. Thanks in advanceIMG_5947.jpg


What I done was divide both sides by 2x2^{x}

So 3=7x−22x3 = \dfrac{7^{x-2}}{2^{x}}

Then take logs of both sides and use the log(xy)=logx−logylog(\frac{x}{y}) = logx - logy rule. Then find a way of factorising xx

Your mistake lies in the first line of your working out. Taking log to 3(2x)3(2^{x}) does not equal 3log(2x)3log(2^{x})

It is log[3(2x)]log[3(2^{x})] and again from here you can use the log(xy)=logx+logylog(xy) = logx + logy rule to work it out also
(edited 8 years ago)
Original post by edothero
What I done was divide both sides by 2x2^{x}

So 3=7x−22x3 = \dfrac{7^{x-2}}{2^{x}}

Then take logs of both sides and use the log(xy)=logx−logylog(\frac{x}{y}) = logx - logy rule. Then find a way of factorising xx

Your mistake lies in the first line of your working out. Taking log to 3(2x)3(2^{x}) does not equal 3log(2x)3log(2^{x})

It is log[3(2x)]log[3(2^{x})] and again from here you can use the log(xy)=logx+logylog(xy) = logx + logy rule to work it out also


THANK YOU SO MUCH! Everything became a lot clearer once I divided by 2^x
Original post by doglover123
THANK YOU SO MUCH! Everything became a lot clearer once I divided by 2^x


No problem I'm glad I helped. Also I hope you took note about your mistake and why it was wrong! Always good to learn from mistakes :smile:
(edited 8 years ago)

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