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!