Log3(x)=log9(6x)

1. 👍 0
2. 👎 0
3. 👁 579
1. since 9 = 3^2,
log3(n) = 2log9(n)
That is, the power of 3 needed for a number is twice as big as the power of 9.

So, we have

2log9(x) = log9(6x)
log9(x^2) = log9(6x)
x^2 = 6x
x = 0 or 6
0 not allowed, so x=6

1. 👍 0
2. 👎 0

## Similar Questions

1. ### Math

Given that x=log3^5 and y=log3^2, rewrite log3^60 in terms of x and y.

2. ### maths

express log9 in terms of log3 x and log3 y and solve for x and y simultaneously

3. ### Math

Simplify (Log75+log9+log5)÷(log5+log45)

4. ### Algebra

log3 (2x – 1) – log3 (x – 4) = 2

1. ### mathematics

log3(4x+1)-log3(3x-5)=2

2. ### calculus

1) Solve for x in terms of k: log3(x) + log3(x+7) = k; 3 is the base 2) e^(x+5) = e^(x) + 6, solve for x. Please help, I have no idea how to do these and I've tried several times!

3. ### Pre Calculus

Evaluate the function f(x) = log3 x at x =1/27 without using a calculator a. –4 b. –2 c. 1/27 d. –3 e. 27

4. ### Mathematics

Find x such that log3(2x-1)=1+log2(x+1)?

1. ### Maths

Log384÷5+log81÷32+3log5÷3-log9=2

2. ### Algebra 1

Explain the differences between solving these two equations: •log3(x - 1) = 4 AND log3(x - 1) = log34 I think it has to deal with the formulas?

3. ### Algebra II

Write the equation log243(81)=4/5 in exponential form. My answer is 243^4/5=81. Evaluate 9^log9(54). My answer is 6. Are these correct?? Thanks

4. ### algebra

(2*log7 16)/(log3(√10+1)+log3(√10-1)log7 2