![Given that log2a = S, log4b = S^2 and logc^2 (8) = 2s^3+ 1, write log2 a^2b^5c^4 as a function of 'S' ( a,b,c > 0,c 1 ). Given that log2a = S, log4b = S^2 and logc^2 (8) = 2s^3+ 1, write log2 a^2b^5c^4 as a function of 'S' ( a,b,c > 0,c 1 ).](https://haygot.s3.amazonaws.com/questions/2039521_1295973_ans_3c02d6eb0b81489d91b5e8444b7363c4.jpg)
Given that log2a = S, log4b = S^2 and logc^2 (8) = 2s^3+ 1, write log2 a^2b^5c^4 as a function of 'S' ( a,b,c > 0,c 1 ).
SOLUTION: My son submitted the following 2 questions earlier: log base 2 square root 8 and 5^1-2x=1/5 We have the solutions for these but no explanation. Neither of us can figure eith
![SOLVED: Suppose log2(8)=a and log2(6)=b Use the change of base formula along with properties of logarithms to rewrite the following in terms of aa and bb. log6(2)= log6(8/6)= SOLVED: Suppose log2(8)=a and log2(6)=b Use the change of base formula along with properties of logarithms to rewrite the following in terms of aa and bb. log6(2)= log6(8/6)=](https://cdn.numerade.com/ask_previews/9294ad38-50af-49f1-bb71-01f168a08588_large.jpg)
SOLVED: Suppose log2(8)=a and log2(6)=b Use the change of base formula along with properties of logarithms to rewrite the following in terms of aa and bb. log6(2)= log6(8/6)=
![log to the base 8 log to the base 4 log to the base 2 16 x then the value of x is what jcgnxb44 -Maths - TopperLearning.com log to the base 8 log to the base 4 log to the base 2 16 x then the value of x is what jcgnxb44 -Maths - TopperLearning.com](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=4b7754f387437a6e5e67f07cd965611a.png&d=0)