Can the solution of exponential equation be expressed in logarithmic form? For example, can the solution of 9 ^ X-6 * 3 ^ X-7 = 0 be expressed as log3 7 (3 is the base)?

Can the solution of exponential equation be expressed in logarithmic form? For example, can the solution of 9 ^ X-6 * 3 ^ X-7 = 0 be expressed as log3 7 (3 is the base)?

Of course, we can use the exchange element to find 3 ^ x, which is 7
Find x again

Solving the equation about logarithm
log(12)[x^(1/2)+x^(1/4)]=[log(9)x]/2

I'm sorry. I always make mistakes. No, carelessness is an ability. I'm a little lacking in this ability
Let x ^ (1 / 4) = Y > 0, and the equation is reduced to
log(12)[y^2 + y] = log(9)y^2 = log(3)y
Let's use the formula e as the base
ln(y^2+y)/ln12 = lny/ln3
[lny + ln(y+1)]/lny = ln12/ln3
1 + ln(y+1)/lny = 1 + ln4/ln3
ln(y+1)/lny = ln4/ln3
LNY, ln (y + 1) and ln (y + 1) / LNY are all one-to-one mapping functions
So y = 3
x = 3^4 = 81

On the high one of logarithmic exponential equation
X power of 4 + x power of 4 - 6 [x power of 2 + x power of 2] + 10 = 0
Write about the process. If you can do several, I will add 100
Oh, and
6^（2x+4）=3^3 *2^(x+8)
3* 4^x + 2*9^x = 5* 6^x
If the domain of [log (x-3)] is [4,11], then the domain of F (x) is?

4 ^ x + 4 ^ (- x) - 6 (2 ^ x + 2 ^ (- x)) + 10 = 0 {(2 ^ x + 2 ^ (- x))} ^ 2-6 (2 ^ x + 2 ^ (- x)) + 8 = 0 let 2 ^ x + 2 ^ (- x) = t, then T ^ 2-6t + 8 = 0 get t = 2 or T = 4, that is, 2 ^ x + 2 ^ (- x) = 2 or 4 get 2 ^ x = 1 or 2 ^ x = 2 ± √ 3 x = 0 or X = log_ 2^(2±√3)...