Question

1. indices, logarithms

exponential

1. solve the equations

a = 3x = 6
b = 4^{2x+1} = 3
c = (5^x)(5x^{-4}) = 10
(d) = 2^{2x} = 5 ; e = 2(2^{2x}) - 5(2^x) + 2 = 0
f = 3^{2x+1} - 26(3^x) - 9 = 0
g = 4^x - 6(2^x) - 16 = 0
h = log_2 7 + log_{1/4} 1/2 = log_9 x
l = log_2 8^{x^3} - log_x 9^{x^2} = 8

Explanation:

Step1: Solve \(3^x = 6\)

Take the natural logarithm of both sides: \(\ln(3^x)=\ln(6)\).
Using the property \(\ln(a^b)=b\ln(a)\), we get \(x\ln(3)=\ln(6)\).
Then \(x = \frac{\ln(6)}{\ln(3)}\).

Step2: Solve \(4^{2x + 1}=3\)

Rewrite \(4\) as \(2^2\), so \((2^2)^{2x+1}=3\), which simplifies to \(2^{4x + 2}=3\).
Take the natural logarithm: \((4x + 2)\ln(2)=\ln(3)\).
Solve for \(x\): \(4x+2=\frac{\ln(3)}{\ln(2)}\), \(4x=\frac{\ln(3)}{\ln(2)}- 2\), \(x=\frac{\frac{\ln(3)}{\ln(2)}-2}{4}\).

Step3: Solve \((5^x)(5^{x - 4})=10\)

Using the property \(a^m\times a^n=a^{m + n}\), we have \(5^{2x-4}=10\).
Take the natural logarithm: \((2x - 4)\ln(5)=\ln(10)\).
Solve for \(x\): \(2x-4=\frac{\ln(10)}{\ln(5)}\), \(2x=\frac{\ln(10)}{\ln(5)}+4\), \(x=\frac{\frac{\ln(10)}{\ln(5)}+4}{2}\).

Step4: Solve \(2^{2x}=5\)

Take the natural logarithm: \(2x\ln(2)=\ln(5)\).
Solve for \(x\): \(x=\frac{\ln(5)}{2\ln(2)}\).

Step5: Solve \(2(2^{2x})-5(2^x)+2 = 0\)

Let \(y = 2^x\), then the equation becomes \(2y^2-5y + 2=0\).
Factor: \((2y - 1)(y - 2)=0\).
So \(y=\frac{1}{2}\) or \(y = 2\).
If \(y = 2^x=\frac{1}{2}\), then \(x=-1\); if \(y = 2^x=2\), then \(x = 1\).

Step6: Solve \(3^{2x+1}-26(3^x)-9 = 0\)

Let \(y = 3^x\), then \(3y^2-26y - 9=0\).
Using the quadratic formula \(y=\frac{26\pm\sqrt{26^2-4\times3\times(-9)}}{2\times3}=\frac{26\pm\sqrt{676 + 108}}{6}=\frac{26\pm\sqrt{784}}{6}=\frac{26\pm28}{6}\).
We get \(y = 9\) or \(y=-\frac{1}{3}\) (reject \(y=-\frac{1}{3}\) since \(3^x>0\)).
If \(y = 3^x=9\), then \(x = 2\).

Step7: Solve \(4^x-6(2^x)-16 = 0\)

Rewrite \(4^x=(2^x)^2\), let \(y = 2^x\), then \(y^2-6y - 16=0\).
Factor: \((y - 8)(y+2)=0\).
So \(y = 8\) or \(y=-2\) (reject \(y=-2\)).
If \(y = 2^x=8\), then \(x = 3\).

Step8: Solve \(\log_21+\log_{4}^{\frac{1}{2}}=\log_9x\)

Since \(\log_21 = 0\), and \(\log_{4}^{\frac{1}{2}}=\frac{\ln(\frac{1}{2})}{\ln(4)}=-\frac{1}{2}\).
Then \(-\frac{1}{2}=\log_9x\), so \(x=9^{-\frac{1}{2}}=\frac{1}{3}\).

Step9: Solve \(\log_2^{8x^3}-\log_x^{8x^5}=8\)

Using the change - of - base formula \(\log_a b=\frac{\ln b}{\ln a}\).
\(\log_2^{8x^3}=\frac{\ln(8x^3)}{\ln(2)}=\frac{\ln8 + 3\ln x}{\ln2}=3 + \frac{3\ln x}{\ln2}\), \(\log_x^{8x^5}=\frac{\ln(8x^5)}{\ln x}=\frac{\ln8+5\ln x}{\ln x}=\frac{3\ln2 + 5\ln x}{\ln x}=\frac{3\ln2}{\ln x}+5\).
Let \(t=\frac{\ln x}{\ln2}\), then the equation becomes \(3 + 3t-( \frac{3}{t}+5)=8\).
Multiply through by \(t\): \(3t+3t^2-3 - 5t=8t\).
Rearrange: \(3t^2-10t - 3=0\).
Using the quadratic formula \(t=\frac{10\pm\sqrt{100 + 36}}{6}=\frac{10\pm\sqrt{136}}{6}=\frac{10\pm2\sqrt{34}}{6}=\frac{5\pm\sqrt{34}}{3}\).
Then \(x = 2^t=2^{\frac{5\pm\sqrt{34}}{3}}\).

Answer:

a. \(x=\frac{\ln(6)}{\ln(3)}\)
b. \(x=\frac{\frac{\ln(3)}{\ln(2)}-2}{4}\)
c. \(x=\frac{\frac{\ln(10)}{\ln(5)}+4}{2}\)
d. \(x=\frac{\ln(5)}{2\ln(2)}\)
e. \(x=-1\) or \(x = 1\)
f. \(x = 2\)
g. \(x = 3\)
h. \(x=\frac{1}{3}\)
i. \(x = 2^{\frac{5\pm\sqrt{34}}{3}}\)