Solve for y in each of the following logarithmic equations:
\log_{7} y = 5
\log_{16} y = \dfrac{1}{2}
\log_{y} 8 = 3
2\log_{2} y + 3\log_{2} y = 15
\log_{6} y^2 = 4
\log_{2} y^{4} + \log_{2} y = 10
Solve for x in each of the following logarithmic equations:
\log_{2} 16 = x
\log_{12} 144 = x
\log_{3} x = 4
2 \log x = 4
\log_{0.5} x = 10
\log_{16} x = \dfrac{1}{4}
\log_{x} 64 = 2
\log_{x} 125 = 3
\log_{9} \left( 7 x + 5\right) = 4
\log_{6} \left( 3 x - 9\right) = 2
\log_{10} \left( 3 x + 982\right) = 3
6 \log_{4} \left( 2 x\right) - 18 = 0
11 \log_{5} \left(x - 12\right) = 33
\log_{2} \left(\sqrt{ 2 x^{3}}\right) + 1 = 4.5
\log_{2} \left(\log_{2} x\right) = 0
\log_{2} \left( 3 x + 35\right) = \log_{2} 5
\log_{9} \left( 10 x - 3\right) = \log_{9} \left( 5 x + 4\right)
Solve for x in each of the following logarithmic equations:
\log_{7} \left( 9 x - 2\right) = \log_{7} \left( 8 x + 9\right)
\log_{6} x = \sqrt{\log_{6} x}
\log_{2} x - 4 = \sqrt{\log_{2} x - 4}
3 \log x = \log 125
\log_{10} x + \log_{10} 6 = \log_{10} 48
\log_{7} 2 x + \log_{7} 3 = 3
\log_{4} 45 - \log_{4} 9 x = 2
\log_{4} x = \log_{4} 7 - \log_{4} \left(x - 4\right)
\log x + \log \left(x + 3\right) = 1
\log_{3} x + \log_{3} 25 x = 8
\log_{5} \left( 4 x^{2}\right) - \log_{5} x = 3
\log \left(x + 12\right) - \log \left(x + 5\right) = \log x
\log_{8} \left( 5 x + 12\right) = \log_{8} \left(x + 6\right) - \log_{8} 3
\log \left(x + 5\right) + \log \left(x - 2\right) = \log 8
\log_{4} \left(x + 3\right) + \log_{4} \left(x - 3\right) = 2
Find the interval in which the solution of the following equations will lie:
5^{x} = 12
3^{x} = 29
2^{x} = \dfrac{1}{13}
For each of the following equations:
Make x the subject of the equation.
Evaluate x to three decimal places.
3^{x} = 11
11^{x} = 3
7^{x} = \dfrac{1}{13}
343^{x} = 12\,167
7^{ - x } = 21
4^{x - 2} = 10
4^{ 2 x - 8} = 70
For each of the following incorrect sets of working:
Which step was incorrect? Explain your answer.
Evaluate a to three decimal places.
\begin{aligned} 9 ^ {a} &= 40 \\ \log9^{a} &= \log40 & (1)\\ a + \log 9 &= \log40 & (2)\\ a &= \log40 - \log9 & (3)\\ &\approx 0.648 & (4) \end{aligned}
\begin{aligned} 2 ^ {a} &= 89 \\ \log2^{a} &= \log89 & (1) \\ a \log 2 &= \log89 & (2)\\ a &= \log_{89} 2 & (3)\\ &\approx 0.154 & (4) \end{aligned}
Solve for y in each of the following equations, correct to two decimal places:
8^{y} = 15
2^{y + 1} = 11
3^{y + \left( - 5 \right)} = 11^{y}
5^{y + \left( - 4 \right)} = 6^{y + 3}
Solve for the exact value of x in each of the following equations:
2 \left(11^{x}\right) = 4
\dfrac{1}{3} \left(11^{x}\right) = 2
2^{ 3 x} = 3^{x - 1}
5^{ 5 x} = 3^{ 2 x + 3}
5^{x} = 2^{ 4 x + 6}
2^{ 3 x} = 3^{ 7 x - 1}
2^{ 3 x + 4} = 5^{x - 1}
2^{ 2 x - 7} = 7^{ 3 x + 5}
5 \left(2^{x}\right) = 7 - 2^{x + 3}
\left(7^{x}\right)^{2} - 4 \left(7^{x}\right) = 0
\left(5^{x}\right)^{2} + 2 \left(5^{x}\right) - 24 = 0
\left(3^{x}\right)^{2} - 3 \left(3^{x}\right) = 10
Solve for x in each of the following equations, correct to two decimal places:
2^{6 - x} = \dfrac{1}{14}
3^{x} + 3 = 84
\left(\dfrac{1}{6}\right)^{x + 3} = \sqrt{7}
7^{x} = 3^{x + 1}
3 \left(5^{ 2 x - 1}\right) + 1 = 5
2 \log_{10}^{2} x - \log_{10} x = 6
\log_{10} x = x^{2} - 12 x + 8
2^{x} = 3 - 7^{x + 1}
2^{x} = \log_{10} \left(x + 5\right)
x = 6^{x}
\left(\log_{5} x\right)^{2} - \log_{5} \left(x^{6}\right) + 9 = 0
3 x + 2 = 4^{x}
The equation for the population at time t is given by Q = 30 \times 8^{ 5 t}. Make t the subject of the equation.
If A = 5 \log y - 50, make y the subject of the equation.
Rearrange the following equations to make M the subject:
u = 5 \log \left(\dfrac{M}{N}\right)
m = 2.1 \log \left(\dfrac{M}{N}\right) + 8
Given that \log p = \log q + r \log x, form an equation in which p is the subject, and no logarithms are involved.
If A = P \left(\dfrac{1 - \left(1 + r\right)^{ - t }}{r}\right), make t the subject of the equation.