6x286x3Solution6x 286x3 Let 6x y Then we have y28y 3 On mu

6^x-28*6^-x=3

Solution

6^x -28*6^-x=3.

Let 6^x = y. Then, we have y-28/y = 3. On multiplying both the sides by y, we get y²-28=3y or, y²- 3y -28 =0 or, y²- 7y+4y -28 = 0 or, y(y-7)+4(y-y) = 0 or, (y-7)(y+4) = 0. Hence, either y = 7 or y = -4. Thus, either 6^x = 7 or, 6^x = -4. Now, if 6^x = 7, then, on taking logarithm of both the sides, we have x log 6 = log 7, so that x = log 7/log6. Similarly, if 6^x = -4, then on taking logarithm of both the sides, we have x log 6 = log (-4), so that x = log 7/log(-4). Now, since the logarithm of a negative real number is undefined, the only real solution is x = log 7/log6.