Question

How to solve complicated logarithmic equations?

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

Answer 1

`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`

Answer 2

`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!