How would I solve this problem - LOGS with different bases? 10 points?

Change of base formula: log_a(b) = log_c(b) / log_c(a)

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

(2x - 1)*ln(4) = (x + 1)*ln(3)

2ln(4)x - ln(4) = ln(3)x + ln(3)

(2ln(4) - ln(3))x = ln(3) + ln(4)

ln(16/3)*x = ln(12)

x = ln(12) / ln(16/3)

x ~ 1.48443347

log_3(x) = log_9(7x - 6)

9^(log_3(x)) = 9^(log_9(7x - 6))

(3^2)^(log_3(x)) = 7x - 6

3^(2*log_3(x)) = 7x - 6

3^(log_3(x²)) = 7x - 6

x² = 7x - 6

x² - 7x + 6 = 0

(x - 1)(x - 6) = 0

x = 1, 6

Both answers work, but you must make sure a negative doesn't result from your answers.

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

x = 1.48443

Log[3, x] =Log[9, 7 x - 6]

x = 1},

{x = 6}

The organic log is the in maximum circumstances used. in case you have been to get used to utilising log base 10, you will be able to ought to apply substitute in base formulation faster or later. to circumvent this, attempt getting used to utilising the organic log. In a feeling, there's a small volume of advantage in his declare, yet once you tell a professor which you took the log, he will understand what you propose regardless, as long as you persist with the simple regulations of algebra.

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

(2x - 1) = (x + 1)*ln(3)/ln(4)

Now solve for x.