How to solve complicated logarithmic equations?

enter image source here

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

2 Answers
Jul 9, 2017

x=5/3 and x=2/3

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 = (log_10)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 = log_10(2x + 1) - 2log_10(x + 1) + log_10(5x + 8)

Use log_a n + log_a m = log_a(nm) and log_a n - log_a m = log_a(n/m) and blog_a n = log_a n^b

1 = log_10 (((2x + 1)(5x+ 8))/(x + 1)^2)

1 = log_10( (10x^2 + 21x + 8)/(x + 1)^2)

Convert to exponential form. If a = log_b c, then c = b^a.

10^1 = (10x^2 + 21x + 8)/(x + 1)

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

10x^2 + 20x + 10 = 10x^2 + 21x + 8

10 - 8 = 21x - 20x

x = 2

Hopefully this helps!