Logarithmic equations ranging from a to v, including problems with different bases and algebraic manipulations.

Here are the step-by-step solutions for each logarithmic equation.

a) $\log_3(1-x) = \log_3(x+16-x^2)$
Since the bases are the same, we can equate the arguments:
$1 - x = x + 16 - x^2$
Rearrange into a quadratic equation:
$x^2 - 2x - 15 = 0$
Factor the quadratic:
$(x - 5)(x + 3) = 0$
Possible solutions: $x = 5$ or $x = -3$.
Check validity (arguments must be positive):
For $x=5$: $\log_3(1-5) = \log_3(-4)$, which is undefined.
For $x=-3$: $\log_3(1-(-3)) = \log_3(4)$ and $\log_3(-3+16-9) = \log_3(4)$. This works.
Solution: $x = -3$

b) $3\log 2 - \log(x-1) = \log(x+1) - \log(x-2)$
Assume base 10. Combine logs on each side:
$\log(2^3) - \log(x-1) = \log(\frac{x+1}{x-2})$
$\log(\frac{8}{x-1}) = \log(\frac{x+1}{x-2})$
Equate arguments:
$\frac{8}{x-1} = \frac{x+1}{x-2}$
Cross-multiply:
$8(x-2) = (x-1)(x+1)$
$8x - 16 = x^2 - 1$
$x^2 - 8x + 15 = 0$
$(x-3)(x-5) = 0$
Possible solutions: $x=3, x=5$.
Check validity ($x>2$ for $\log(x-2)$ to exist): Both 3 and 5 are greater than 2.
Solutions: $x = 3, x = 5$

c) $\frac{\log x}{\log x + 1} = -1$
Multiply both sides by $(\log x + 1)$:
$\log x = -1(\log x + 1)$
$\log x = -\log x - 1$
$2\log x = -1$
$\log x = -0.5$
$x = 10^{-0.5} = \frac{1}{\sqrt{10}}$
Solution: $x = \frac{1}{\sqrt{10}}$

d) $\frac{\log_2(9-2^x)}{3-x} = 1$
$\log_2(9-2^x) = 3-x$
Convert to exponential form:
$9 - 2^x = 2^{3-x}$
$9 - 2^x = \frac{2^3}{2^x} = \frac{8}{2^x}$
Let $u = 2^x$. Then $9 - u = \frac{8}{u}$.
$9u - u^2 = 8 \Rightarrow u^2 - 9u + 8 = 0$
$(u-8)(u-1) = 0$
So $u=8$ or $u=1$.
If $2^x = 8$, then $x=3$. But the original denominator is $3-x$, so $x \neq 3$.
If $2^x = 1$, then $x=0$. Denominator is $3-0=3 \neq 0$. Valid.
Solution: $x = 0$

e) $\log_3[1+\log_3(2^x-7)] = 1$
$1 + \log_3(2^x-7) = 3^1 = 3$
$\log_3(2^x-7) = 2$
$2^x - 7 = 3^2 = 9$
$2^x = 16$
$x = 4$
Solution: $x = 4$

f) $\log_{5-x}(x^2-2x+65) = 2$
Definition of log: $(5-x)^2 = x^2 - 2x + 65$
Constraints: Base $5-x > 0, 5-x \neq 1 \Rightarrow x < 5, x \neq 4$. Argument $> 0$.
Expand:
$25 - 10x + x^2 = x^2 - 2x + 65$
Subtract $x^2$ from both sides:
$25 - 10x = -2x + 65$
$-8x = 40$
$x = -5$
Check constraints: $x=-5 < 5$ and $x \neq 4$. Base is $5-(-5)=10$. Argument is $25+10+65=100$. $\log_{10}100=2$. Valid.
Solution: $x = -5$

g) $(\log_3 x)^2 - \log_3 x^3 + 2 = 0$
Use property $\log_3 x^3 = 3\log_3 x$. Let $u = \log_3 x$.
$u^2 - 3u + 2 = 0$
$(u-2)(u-1) = 0$
$u=2 \Rightarrow \log_3 x = 2 \Rightarrow x = 3^2 = 9$
$u=1 \Rightarrow \log_3 x = 1 \Rightarrow x = 3^1 = 3$
Solutions: $x = 3, x = 9$

h) $\log x + \frac{1}{\log x} = 2$
Let $u = \log x$.
$u + \frac{1}{u} = 2$
Multiply by $u$: $u^2 + 1 = 2u \Rightarrow u^2 - 2u + 1 = 0$
$(u-1)^2 = 0 \Rightarrow u = 1$
$\log x = 1 \Rightarrow x = 10^1 = 10$
Solution: $x = 10$

i) $\log_7 x + \log_x 7 = 2.5$
Note that $\log_x 7 = \frac{1}{\log_7 x}$. Let $u = \log_7 x$.
$u + \frac{1}{u} = 2.5 = \frac{5}{2}$
$2u^2 + 2 = 5u \Rightarrow 2u^2 - 5u + 2 = 0$
$(2u-1)(u-2) = 0$
$u = 1/2 \Rightarrow \log_7 x = 1/2 \Rightarrow x = 7^{1/2} = \sqrt{7}$
$u = 2 \Rightarrow \log_7 x = 2 \Rightarrow x = 7^2 = 49$
Solutions: $x = \sqrt{7}, x = 49$

j) $\log_7(2^x-1) + \log_7(2^x-7) = 1$
Combine logs: $\log_7[(2^x-1)(2^x-7)] = 1$
$(2^x-1)(2^x-7) = 7^1 = 7$
Let $u = 2^x$. Note constraints: $2^x > 1$ and $2^x > 7 \Rightarrow u > 7$.
$(u-1)(u-7) = 7$
$u^2 - 8u + 7 = 7$
$u^2 - 8u = 0 \Rightarrow u(u-8) = 0$
$u=0$ (invalid since $2^x>0$) or $u=8$.
$2^x = 8 \Rightarrow x = 3$.
Check constraint: $2^3 = 8 > 7$. Valid.
Solution: $x = 3$

k) $x^{1+\log x} = 10x$
Take $\log$ (base 10) of both sides:
$\log(x^{1+\log x}) = \log(10x)$
$(1+\log x)\log x = \log 10 + \log x$
Let $u = \log x$.
$(1+u)u = 1 + u$
$u + u^2 = 1 + u$
$u^2 = 1 \Rightarrow u = 1$ or $u = -1$
If $u=1, \log x = 1 \Rightarrow x=10$.
If $u=-1, \log x = -1 \Rightarrow x=0.1$.
Solutions: $x = 10, x = 0.1$

l) $\log^2 x - 3\log x = \log x^2 - 4$
Assume $\log^2 x$ means $(\log x)^2$. And $\log x^2 = 2\log x$.
Let $u = \log x$.
$u^2 - 3u = 2u - 4$
$u^2 - 5u + 4 = 0$
$(u-4)(u-1) = 0$
$u=4 \Rightarrow x=10^4=10000$
$u=1 \Rightarrow x=10^1=10$
Solutions: $x = 10, x = 10000$

m) $\log_2(4 \cdot 3^x - 6) - \log_2(9^x - 6) = 1$
$\log_2(\frac{4 \cdot 3^x - 6}{9^x - 6}) = 1$
$\frac{4 \cdot 3^x - 6}{9^x - 6} = 2^1 = 2$
Let $u = 3^x$. Then $9^x = u^2$.
$\frac{4u - 6}{u^2 - 6} = 2$
$4u - 6 = 2(u^2 - 6)$
$4u - 6 = 2u^2 - 12$
$2u^2 - 4u - 6 = 0 \Rightarrow u^2 - 2u - 3 = 0$
$(u-3)(u+1) = 0$
$u=3$ or $u=-1$. Since $u=3^x > 0$, only $u=3$ works.
$3^x = 3 \Rightarrow x=1$.
Check args: $4(3)-6=6>0$, $9-6=3>0$. Valid.
Solution: $x = 1$

n) $\frac{\log(35-x^3)}{\log(5-x)} = 3$
This is equivalent to $\log_{5-x}(35-x^3) = 3$.
$(5-x)^3 = 35 - x^3$
$125 - 75x + 15x^2 - x^3 = 35 - x^3$
$15x^2 - 75x + 90 = 0$
Divide by 15: $x^2 - 5x + 6 = 0$
$(x-2)(x-3) = 0$
$x=2$ or $x=3$.
Check base $5-x$: Must be $>0$ and $\neq 1$.
If $x=2$, base $= 3$. OK.
If $x=3$, base $= 2$. OK.
Check arguments:
$x=2: 35-8=27 > 0$. OK.
$x=3: 35-27=8 > 0$. OK.
Solutions: $x = 2, x = 3$

o) $\log_{16} x + \log_4 x + \log_2 x = 7$
Change all to base 2:
$\log_{16} x = \frac{\log_2 x}{\log_2 16} = \frac{\log_2 x}{4}$
$\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}$
Let $u = \log_2 x$.
$\frac{u}{4} + \frac{u}{2} + u = 7$
Multiply by 4: $u + 2u + 4u = 28$
$7u = 28 \Rightarrow u = 4$
$\log_2 x = 4 \Rightarrow x = 2^4 = 16$
Solution: $x = 16$

p) $\log(64 \sqrt[24]{2^{x^2-40x}}) = 0$
Argument must equal $10^0 = 1$ (assuming base 10) or $e^0=1$. In any base $b$, $\log_b A = 0 \implies A=1$.
$64 \cdot (2^{x^2-40x})^{1/24} = 1$
$2^6 \cdot 2^{\frac{x^2-40x}{24}} = 1 = 2^0$
Exponents add up:
$6 + \frac{x^2-40x}{24} = 0$
$\frac{x^2-40x}{24} = -6$
$x^2 - 40x = -144$
$x^2 - 40x + 144 = 0$
Roots: $x = \frac{40 \pm \sqrt{1600 - 576}}{2} = \frac{40 \pm \sqrt{1024}}{2} = \frac{40 \pm 32}{2}$
$x_1 = \frac{72}{2} = 36$
$x_2 = \frac{8}{2} = 4$
Solutions: $x = 4, x = 36$

q) $3\sqrt{\log x} + 2\log \sqrt{x^{-1}} = 2$
Term 2: $2\log (x^{-1/2}) = 2(-\frac{1}{2}\log x) = -\log x$.
Let $u = \sqrt{\log x}$. Then $\log x = u^2$. Note $u \ge 0$.
$3u - u^2 = 2$
$u^2 - 3u + 2 = 0$
$(u-1)(u-2) = 0$
$u=1 \Rightarrow \sqrt{\log x}=1 \Rightarrow \log x = 1 \Rightarrow x=10$.
$u=2 \Rightarrow \sqrt{\log x}=2 \Rightarrow \log x = 4 \Rightarrow x=10000$.
Solutions: $x = 10, x = 10000$

r) $\log_7 2 + \log_{49} x = \log_{\frac{1}{7}} \sqrt{3}$
Convert to base 7:
$\log_{49} x = \frac{\log_7 x}{\log_7 49} = \frac{\log_7 x}{2}$
$\log_{\frac{1}{7}} \sqrt{3} = \frac{\log_7 \sqrt{3}}{\log_7 7^{-1}} = \frac{\frac{1}{2}\log_7 3}{-1} = -\frac{1}{2}\log_7 3$
Equation:
$\log_7 2 + \frac{1}{2}\log_7 x = -\frac{1}{2}\log_7 3$
Multiply by 2:
$2\log_7 2 + \log_7 x = -\log_7 3$
$\log_7 4 + \log_7 x = \log_7 (3^{-1})$
$\log_7 (4x) = \log_7 (\frac{1}{3})$
$4x = \frac{1}{3} \Rightarrow x = \frac{1}{12}$
Solution: $x = \frac{1}{12}$

s) $\log_4(x+12) \cdot \log_x 2 = 1$
$\frac{\log_2(x+12)}{\log_2 4} \cdot \frac{\log_2 2}{\log_2 x} = 1$
$\frac{\log_2(x+12)}{2} \cdot \frac{1}{\log_2 x} = 1$
$\log_2(x+12) = 2\log_2 x = \log_2 x^2$
$x+12 = x^2$
$x^2 - x - 12 = 0$
$(x-4)(x+3) = 0$
$x=4$ or $x=-3$.
Base $x$ must be $>0, \neq 1$. So $x=-3$ is invalid.
Check $x=4$: $\log_4(16) \cdot \log_4 2 = 2 \cdot 0.5 = 1$. Valid.
Solution: $x = 4$

t) $5^{\log x} = 50 - x^{\log 5}$
Identity: $a^{\log b} = b^{\log a}$.
So $5^{\log x} = x^{\log 5}$.
Let $y = x^{\log 5}$.
$y = 50 - y \Rightarrow 2y = 50 \Rightarrow y = 25$.
$x^{\log 5} = 25$
Take $\log$ of both sides:
$\log(x^{\log 5}) = \log 25$
$\log 5 \cdot \log x = \log 25$
$\log x = \frac{\log 25}{\log 5} = \frac{\log 5^2}{\log 5} = \frac{2\log 5}{\log 5} = 2$
$x = 10^2 = 100$
Solution: $x = 100$

u) $\frac{2\log x}{\log(5x-4)} = 1$
$2\log x = \log(5x-4)$
$\log x^2 = \log(5x-4)$
$x^2 = 5x - 4$
$x^2 - 5x + 4 = 0$
$(x-4)(x-1) = 0$
$x=4$ or $x=1$.
Check validity:
If $x=1$, $\log(5(1)-4) = \log 1 = 0$. Division by zero. Invalid.
If $x=4$, $\log(20-4)=\log 16 \neq 0$. Valid.
Solution: $x = 4$

v) $\log(\log x) + \log(\log x^3 - 2) = 0$
Assume base 10.
$\log(\log x) + \log(3\log x - 2) = 0$
Let $u = \log x$. Constraints: $u>0$ (for first term) and $3u-2>0 \Rightarrow u>2/3$.
$\log u + \log(3u-2) = 0$
$\log[u(3u-2)] = 0$
$u(3u-2) = 10^0 = 1$
$3u^2 - 2u - 1 = 0$
$(3u+1)(u-1) = 0$
$u = -1/3$ (invalid, $u>2/3$) or $u=1$.
$u=1 \Rightarrow \log x = 1 \Rightarrow x=10$.
Check: $\log(\log 10) + \log(\log 1000 - 2) = \log 1 + \log(3-2) = 0 + 0 = 0$. Valid.
Solution: $x = 10$

Final Answer:
a) $x = -3$
b) $x = 3, 5$
c) $x = \frac{1}{\sqrt{10}}$
d) $x = 0$
e) $x = 4$
f) $x = -5$
g) $x = 3, 9$
h) $x = 10$
i) $x = \sqrt{7}, 49$
j) $x = 3$
k) $x = 10, 0.1$
l) $x = 10, 10000$
m) $x = 1$
n) $x = 2, 3$
o) $x = 16$
p) $x = 4, 36$
q) $x = 10, 10000$
r) $x = \frac{1}{12}$
s) $x = 4$
t) $x = 100$
u) $x = 4$
v) $x = 10$