Transcript for:
Solving Logarithmic Equations

[Music] hello welcome to the tutorial session of this week nine so let us see the first question so in the first question we are seeing an equation which is given as three log p 9 this is base p 2 log q base 27 and 2 log here basis p inverse q and this is 81 so here it is also given that p and q are two distinct positive real numbers and we have to show either p equal to q q or q equal to p cube so let us try to solve this equation so here we can see that we can write log 9 as 3 square log 27 as 3 cube and we can write this as 3 power 4 so let us take this powers before this log so it will become 6 log p 3 minus 6 log q is the base this is 3 and we can write 8 here and it is p inverse q and here it is three so we transform everything in terms of log three so uh our motivation is to bring everything as the same base so we can write it as so we can take c common and we can write log 3 base p as log p base 3 here we can write it as 1 by log here base 3 and here it is q and we can write this as also 1 by base is 3 and here it is p inverse q okay so let us see what we can do here so one more step so here 2 cancels up it is 3 and here it is 4 and one more step we can do before going ahead so this is my log this portion remains same and this portion i can write it as so this is p inverse q p inverse q means q by p so here it is given uh distinct positive real numbers so it is non-zero so this is the p inverse is valid basically and here also see that uh p and q cannot be negative also otherwise this log will be undefined negative or zero so otherwise this log will be undefined so we are taking two distinct positive real numbers so s p and q so here it is 1 by log 3 q minus log 3 p yeah so we have written every term as in terms of log 3 p or log 3 q so let us assume log p base 3 as m and log cube is 3 as n so what we get if we substitute these things we will get this is 1 by m 1 by n 4 1 by n minus m okay so let us see what this gives so this will give n minus m by m n 4 1 by n minus m so if we take n minus m in this side we will get n minus m whole square equal to 4 mn so we if we simplify this thing we will get 3 let us break it down and square plus m square minus 2 mn here it is 4 mn so if we take everything in the same side we will get 3 n square minus 3 into 2 6 and here it is minus 4 11 so 10 m n plus 3 n square so from here we can write it as m minus 3 n into 3 m minus n equal to 0 so if you factorize this thing you will get this term because 3 into 3 is 9 and we can write this minus 10 as minus 9 minus 1 so you can simplify this thing so basically we get m equal to 3n or 3 m equals to n so whatever m and n are so let us substitute here so log p base 3 is equal to 3 n so 3 log 3 sq so it is giving us both side we have log 3 base and it is q so log is an 1 1 function so we get p equal to q cube and again if you substitute in the other term yeah which is 3 m equal to n we will get q equal to p q so this proves that for the given equation what we have started with we get either p equal to q q or q equal to p q