ok ok never mind looking at the related questions.
7^(2x-1) = 7^(2x)/7
4*7^(x-1) = 4*7^x/7
so, doing what they suggested, you have
u^2/7 + 4u/7 - 1/14 = 0
Think you can handle that?
By making the substitution u=7^x and using part a, derive a hidden quadratic equation
7^(2x-1) +4×7^(x-1) -1/14 =0
and hence find all of the solutions to the above equation.
2 answers
7 ^ ( 2 x - 1 ) = 7 ^ ( 2 x ) / 7 ^ 1 = ( 7 ^ x ) ^ 2 / 7
7 ^ ( x - 1 ) = 7 ^ x / 7 ^ 1 = 7 ^ x / 7
7 ^ ( 2 x -1 ) + 4 * 7 ^ ( x - 1 ) - 1 / 14 = 0
( 7 ^ x ) ^ 2 / 7 + 4 * 7 ^ x / 7 - 1 / 14 = 0
Make the substitution
u = 7 ^ x
u ^ 2 / 7 + 4 u / 7 - 1 / 14 = 0 Multiply both sides by 14
14 u ^ 2 / 7 + 4 * 14 u / 7 - 14 / 14 = 0
14 u ^ 2 / 7 + 56 u / 7 - 1 = 0
2 u ^ 2 + 8 u - 1 = 0
Now you have quadratic equatin with coefficients:
a = 2, b = 8, c = - 1
u1/2 = [ - b ± sqrooot ( b ^ 2 - 4 ac ) ] / 2 a
u1/2 = [ - 8 ± sqrooot ( 8 ^ 2 - 4 * 2 * ( - 1 ) ) ] / 2 * 2
u1/2 = [ - 8 ± sqrooot ( 64 - 8 * ( - 1 ) ] / 4
u1/2 = [ - 8 ± sqrooot ( 64 + 8 ) ] / 4
u1/2 = [ - 8 ± sqrooot ( 72 ) ] / 4
u1/2 = [ - 8 ± sqrooot ( 72 ) ] / 4
u1/2 = [ - 8 ± sqrooot ( 4 * 18 ) ] / 4
u1/2 = [ - 8 ± sqrooot ( 4 ) * sqroot (18 ) ] / 4
u1/2 = [ - 8 ± 2 * sqroot ( 18 ) ] / 4
u1/2 = [ - 2 * 4 ± 2 * sqroot ( 18 ) ] / 2 * 2
u1/2 = 2 * [ - 4 ± sqroot ( 18 ) ] / 2 * 2
u1/2 = [ - 4 ± sqroot ( 18 ) ] / 2
u1/2 = [ - 4 ± sqroot ( 9 * 2 ) ] / 2
u1/2 = [ - 4 ± sqroot ( 9 ) * sqroot ( 2 ) ] / 2
u1/2 = [ - 4 ± 3 sqroot ( 2 ) ] / 2
u1 = [ - 4 + 3 sqroot ( 2 ) ] / 2
u1 = - 4 / 2 + 3 sqroot ( 2 ) / 2
u1 = - 2 + 3 sqroot ( 2 ) / 2 = 0.12132
u2 = [ - 4 - 3 sqroot ( 2 ) ] / 2
u2 = - 4 / 2 - 3 sqroot ( 2 ) / 2
u2 = - 2 - 3 sqroot ( 2 ) / 2 = - 4.12132
u = 7 ^ x Take the logarithm of both sides
ln u = x * ln 7 Divide both sides by ln 7
ln u / ln 7 = x
x = ln u / ln 7
The logarithm of negative numbers does not exist so you must discard solution u = - 4.12132
Soution is:
x = ln u / ln 7
x = ln 0.12132 / ln 7 =
x = - 2,1093236 / 1.94591 =
x = - 1.083978
7 ^ ( x - 1 ) = 7 ^ x / 7 ^ 1 = 7 ^ x / 7
7 ^ ( 2 x -1 ) + 4 * 7 ^ ( x - 1 ) - 1 / 14 = 0
( 7 ^ x ) ^ 2 / 7 + 4 * 7 ^ x / 7 - 1 / 14 = 0
Make the substitution
u = 7 ^ x
u ^ 2 / 7 + 4 u / 7 - 1 / 14 = 0 Multiply both sides by 14
14 u ^ 2 / 7 + 4 * 14 u / 7 - 14 / 14 = 0
14 u ^ 2 / 7 + 56 u / 7 - 1 = 0
2 u ^ 2 + 8 u - 1 = 0
Now you have quadratic equatin with coefficients:
a = 2, b = 8, c = - 1
u1/2 = [ - b ± sqrooot ( b ^ 2 - 4 ac ) ] / 2 a
u1/2 = [ - 8 ± sqrooot ( 8 ^ 2 - 4 * 2 * ( - 1 ) ) ] / 2 * 2
u1/2 = [ - 8 ± sqrooot ( 64 - 8 * ( - 1 ) ] / 4
u1/2 = [ - 8 ± sqrooot ( 64 + 8 ) ] / 4
u1/2 = [ - 8 ± sqrooot ( 72 ) ] / 4
u1/2 = [ - 8 ± sqrooot ( 72 ) ] / 4
u1/2 = [ - 8 ± sqrooot ( 4 * 18 ) ] / 4
u1/2 = [ - 8 ± sqrooot ( 4 ) * sqroot (18 ) ] / 4
u1/2 = [ - 8 ± 2 * sqroot ( 18 ) ] / 4
u1/2 = [ - 2 * 4 ± 2 * sqroot ( 18 ) ] / 2 * 2
u1/2 = 2 * [ - 4 ± sqroot ( 18 ) ] / 2 * 2
u1/2 = [ - 4 ± sqroot ( 18 ) ] / 2
u1/2 = [ - 4 ± sqroot ( 9 * 2 ) ] / 2
u1/2 = [ - 4 ± sqroot ( 9 ) * sqroot ( 2 ) ] / 2
u1/2 = [ - 4 ± 3 sqroot ( 2 ) ] / 2
u1 = [ - 4 + 3 sqroot ( 2 ) ] / 2
u1 = - 4 / 2 + 3 sqroot ( 2 ) / 2
u1 = - 2 + 3 sqroot ( 2 ) / 2 = 0.12132
u2 = [ - 4 - 3 sqroot ( 2 ) ] / 2
u2 = - 4 / 2 - 3 sqroot ( 2 ) / 2
u2 = - 2 - 3 sqroot ( 2 ) / 2 = - 4.12132
u = 7 ^ x Take the logarithm of both sides
ln u = x * ln 7 Divide both sides by ln 7
ln u / ln 7 = x
x = ln u / ln 7
The logarithm of negative numbers does not exist so you must discard solution u = - 4.12132
Soution is:
x = ln u / ln 7
x = ln 0.12132 / ln 7 =
x = - 2,1093236 / 1.94591 =
x = - 1.083978
2 answers