I am supposed solve these SYMBOLICALLY but i don't know how since i was never taught in school.

Could someone help me solve these?

1. 7(e^(2x))^3 = 35
2. 7^(x^2-x)=7^6
3. log3(5x)=4

Thank YOU very much!!

I'll do the first one:

7(e^(2x))^3 = 35
Divide each side by 7:
(e^(2x))^3 = 35/7=5
Take cube roots:
e^(2x) = &#8731(5)
Take log to the base e
ln(e^(2x)) = ln&#8731(5)
2x = (1/3)ln5
x = (1/6)ln5

for #2, just equate the exponents, since they have the same base:

7^(x^2-x)=7^6
x^2 - x = 6
x^2 - x - 6 = 0 *factor*
(x-3)(x+2) = 0
x=3 and x=-2

note: check if there is extraneous root by substituting the roots back to the original.

for #3, i interpreted this as log with base 3:
log(5x)/log(3) = 4
log(5x) = 4*log(3)
log(5x) = log(3^4)
5x = 81
x=81/5

so there,, =)

Of course! I can help you solve these symbolic equations step-by-step. Here's how you can approach each equation:

1. 7(e^(2x))^3 = 35
To solve this equation, we need to isolate the exponential term. Let's break it down step by step:
- Divide both sides by 7 to get: (e^(2x))^3 = 5
- Take the cube root of both sides: e^(2x) = 5^(1/3)
- Take the natural logarithm (ln) of both sides to eliminate the exponential: ln(e^(2x)) = ln(5^(1/3))
- Use the property that ln(e^a) = a: 2x = (1/3) * ln(5)
- Finally, divide both sides by 2 to solve for x: x = (1/6) * ln(5)

2. 7^(x^2-x) = 7^6
In this equation, we have the same base on both sides, so we can equate the exponents:
x^2 - x = 6
Rearrange the equation to get a quadratic equation:
x^2 - x - 6 = 0
Factor the quadratic equation or use the quadratic formula to find the roots of x:
(x - 3)(x + 2) = 0
Therefore, x = 3 or x = -2

3. log3(5x) = 4
To solve this logarithmic equation, we need to exponentiate both sides using a base of 3:
3^(log3(5x)) = 3^4
Use the property that exponentiation and logarithm are inverse operations: 5x = 3^4
Simplify the right side: 5x = 81
Divide both sides by 5 to solve for x: x = 81/5 or x = 16.2

Please note that these solutions are provided symbolically. If you need numerical values, you can substitute the base of logarithm to the base of exponentiation and calculate the results using a scientific calculator or computer software.