2xlog3 6+log3(4^x-2)<=2x+1

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To solve this inequality, we need to simplify the left side first:

2xlog3 6 + log3(4^x-2) <= 2x + 1
log3(6^(2x)) + log3(4^x - 2) <= 2x + 1
log3(36^x) + log3(4^x - 2) <= 2x + 1
log3(36^x(4^x - 2)) <= 2x + 1
log3(4^(2x)6^x - 2^(2)) <= 2x + 1
log3(16^x6^x - 4) <= 2x + 1
log3(96^x - 4) <= 2x + 1

Now, rewrite the inequality in exponential form:

3^(2x + 1) >= 96^x - 4
3^(2x + 1) >= (3^2)^x 4 - 4
3^(2x + 1) >= 9^x 4 - 4
3^(2x + 1) >= 4 * 9^x - 4

Now, let a = 9^x:

3^(2x + 1) >= 4a - 4
3^(2x + 1) - 4a >= 4

Now, we can find the values of x that satisfy the inequality. However, further steps may be required to simplify or solve this inequality.