Want to Ace Math? Watch This Now! | Tricky Exponential Equation

[@Quest3669]

Let 7^x = t then solving eqn gives
t^5-49t^2- 48t-48= 0 Then
t= 4= 7^x only real soln.
Hence ; x= log_7( 4)

[@gregevgeni1864]

Let 7^x = t , then the equation, after some algebra, is equivalent to
t⁵ - 49 t² - 48 t - 48 = 0
(t-4)(t²+t+1)(t²+3t+12) = 0 (Horner)
=> only real solution t =4.
So 7^x = 4 => log(7^x) = log4 =>
x • log7 = log4 => x = (log4)/(log7).

[@Shobhamaths]

Let 7^x=t;
t^5-49t^2-49t-48=0
t=4, remaining are complex
x=log4(base 7)

[@RashmiRay-c1y]

Let 7^x=a. Then, The given equation becomes a^5-49a^2-48a-48=0 a+4 is a solution. [a^5-49a^2-48a-48]/(a-4) = a^4+4a^3+16a^2+15a+12. Let us try a^4+4a^3+16a^2+15a+12 = (a^2+pa+1)(a^2+qa+12). This gives, by comparison, p+q=4, 12p+q=15, 13+pq=16 > p=1, q=3. So, a^2+a+1=0 or a^2+3a+12=0. These do not yield real solutions. So, a=4 > x =ln4/ln7.

[@adribber]

1/7^(4x-1) + 1/7^(5x-1) + 1/7^(6x-1) = 1/7^(x+1) + 1/7^(5x+1) + 1/7^(6x+1)
Put, t = 7^x
7/t^4 + 7/t^5 + 7/t^6 - 1/7t - 1/7t^5 - 1/7*t^6 = 0
(7*7t^6/t^4 - 1*7t^6/7t) + (7*7t^6/t^5 - 1*7t^6/7*t^5) + (7*7t^6/t^6 - 1*7t^6/7t^6) = 0
7²*t^2 - t^5 +7²t - t + 7² - 1 = 0
(7²*t^2 +7²*t + 7²) - (t^5 + t + 1) = 0
7²*(t^2 + t + 1) - (t^5 + t^3 + t) + (t^3 - 1) = 0
7²*(t^2 + t + 1) - t{(t^4 - t) + (t^2 + t + 1)} + (t - 1)(t^2 + t + 1) = 0
7²*(t^2 + t + 1) - t^2(t^3 - 1) - t(t^2 + t + 1)} + (t - 1)(t^2 + t + 1) = 0
7²*(t^2 + t + 1) - t^2(t - 1)(t^2 + t + 1) - t(t^2 + t + 1) + (t - 1)(t^2 + t + 1) = 0
(t^2 + t + 1)[7² - t^2(t - 1) - t+ (t - 1)] = 0
t^2 + t + 1 = 0,
D < 0 ---> reject
7² - t^2(t - 1) - t+ (t - 1) = 0
- t^3 + t^2 + 7² - 1 = 0
t^3 - t^2 - 48 = 0
t^2(t - 1) = 48
t^2(t - 1) = 4^2*3
t^2 > t - 1
∴ t = 4 ---> t = 7^x
7^x = 4
log7^x = log4
x*log7 = 2log2
x = 2log2/log7
∴ x = 2log(base 7)2

[@offrampc]

7^x = 4.