How to solve complicated logarithmic equations?

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Hi, can someone please help me with question 10 e and f? Thanks!

2 Answers
Jul 9, 2017

x=53 and x=23

Explanation:

These questions are tricky because you have a constant in between all of those logs. To get around this, turn them into logs so in this case:

1=(log10)10

Once you've done this, youcan use your other log laws to solve the equations:

2log5 becomes log25 and when you add logs, you multiply the brackets:

log25+log(x+1)=log(25(x+1))

Do the same to the other side and you'll get:

log25+log(x+1)=log(25(x+1))=log((10(2x+7))

You can now remove the logs:

25(x+1)=10(2x+7)

x=9

The second question will become:

10(x+1)2=(2x+1)(5x+8)

Expand and collect to leave x=2

Jul 9, 2017

x=2

Explanation:

Here's how I would do question f.

Put all the logarithms to one side.

1=log10(2x+1)2log10(x+1)+log10(5x+8)

Use logan+logam=loga(nm) and loganlogam=loga(nm) and blogan=loganb

1=log10((2x+1)(5x+8)(x+1)2)

1=log10(10x2+21x+8(x+1)2)

Convert to exponential form. If a=logbc, then c=ba.

101=10x2+21x+8x+1

10(x+1)2=10x2+21x+8

10x2+20x+10=10x2+21x+8

108=21x20x

x=2

Hopefully this helps!