Math

posted by .

Can someone solve this showing the steps involved

4^log2(2^log2 5)

Similar Questions

1. 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?
2. 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?
3. 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?
4. Math

Hi! Could someone possibly help me solve this?
5. Math

Hi! Could someone possibly help me solve this?
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?
7. 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 positive …
8. 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 positive …
9. 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 positive …
10. 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) …

More Similar Questions