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) = ∛(5)
Take log to the base e
ln(e^(2x)) = ln∛(5)
2x = (1/3)ln5
x = (1/6)ln5
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!!
2 answers
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,, =)
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,, =)