# Maths

How do we solve for x and y where y=(1/3)log2(x) and 8^(2y-1) = 2(x-4)

From the 1st one,I got x = 2^(3y) & by substituting it in the 2nd resulted 8^(2y-1) = 2((2)^(3y) - 4)

How to proceed?

1. 👍 0
2. 👎 0
3. 👁 61
1. ok so far, but messy
Check if the 2nd equation wasn't
8^(2y-1) = 2^(x-4)

1. 👍 0
2. 👎 0
posted by Reiny
2. Thanks Reiny I solved it

1. 👍 0
2. 👎 0
posted by Ash

## Similar Questions

1. ### Urgent math

i need help with these two homework problems Use the Laws of Logarithms to combine the expression into a single logarithm log2 5 − 5 log2 x + 1/2 log2(x + 1) Solve the logarithmic equation for x log2(x + 2) + log2(x − 1) = 2

asked by Emma on August 3, 2015
2. ### math

solve the equation log2(x+4)-log4x=2 the 2 and 4 are lower than the g This is what I got: log2(x+4)+log2(4^x)=2 log2((x+4)*4^x)=2 4^x(x+4)=4 x=0 is a solution???

asked by anonymous on January 9, 2013
3. ### Math

I don't understand how log2 √(1/2) turned into log2 2^(-1/2). Quote: You will have to know the 3 prime properties of logs 1. logk (AB) = logk A + logk B 2. logk(A/B) = logk A - logk B 3. logk (A^n) = n logk A where k is any

asked by Anonymous on June 24, 2015
4. ### Math

I don't understand how log2 √(1/2) turned into log2 2^(-1/2). Quote: You will have to know the 3 prime properties of logs 1. logk (AB) = logk A + logk B 2. logk(A/B) = logk A - logk B 3. logk (A^n) = n logk A where k is any

asked by Anonymous on June 24, 2015
5. ### Math

I don't understand how log2 √(1/2) turned into log2 2^(-1/2). Quote: You will have to know the 3 prime properties of logs 1. logk (AB) = logk A + logk B 2. logk(A/B) = logk A - logk B 3. logk (A^n) = n logk A where k is any

asked by Anonymous on June 24, 2015
6. ### Math

Hello! Could someone please take a look at the problem below and let me know if I made mistakes in simplifying the given equation? I'm quite certain that I did make a mistake somewhere because the original equation and its

asked by Constantine on May 31, 2015
7. ### Math - Logarithmic

Solve: 2^(5x-6) = 7 My work: log^(5x-6) = log7 5x - 6(log2) = log7 5x = log7 + 6(log2) x = (log7 + log2^6) / 5 And textbook answer: (log7) / (log2) What did I do wrong?

asked by Anonymous on December 13, 2007
8. ### Algebra 2

solve log2(3x-1)-log2(x-1)=log2(x+1) i have absolutely no idea how to solve this. can anyone help me, please?

asked by Kelly on November 9, 2010
9. ### Logarithms

I'm working on logarithmic equations and I'm stuck on how my book arrives at the next step. First, they use the change of base formula on, log(sqrt(2))(x^3 - 2) (sqrt(2)) is the base,changing to base 2 log(sqrt(2))(x^3 - 2)=

asked by Helper on March 9, 2011
10. ### Math

Hi! Could someone possibly help me solve this?; log2(x+4)=log2(x^2+2). For my answer, I got "All Real Numbers" or x=0 but I'm not sure if that's right. Thanks a bunch! :)

asked by Sally on November 11, 2014
11. ### Math

Hi! Could someone possibly help me solve this?; log2(x+4)=log2(x^2+2). For my answer, I got "All Real Numbers" or x=0 but I'm not sure if that's right. Thanks a bunch!

asked by Sally on November 12, 2014

More Similar Questions