Question

solve each of the following equations
part a
$12^{x-1}=400$
answer this question on paper.
part b
$log_{4}(3x + 4)^3=15$
answer this question on paper.
part c
$log_{12}x+log_{12}(x + 1)=1$
answer this question on paper.

Explanation:

Part A

Step1: Take log base 12 of both sides

$\log_{12}(12^{x-1}) = \log_{12}400$

Step2: Simplify left side using log rule

$x-1 = \log_{12}400$

Step3: Isolate x

$x = \log_{12}400 + 1$

Step4: Rewrite for decimal form (optional)

$x = \frac{\ln 400}{\ln 12} + 1 \approx \frac{5.9915}{2.4849} + 1 \approx 2.411 + 1 = 3.411$

Part B

Step1: Rewrite log in exponential form

$(3x+4)^3 = 4^{15}$

Step2: Take cube root of both sides

$3x+4 = 4^{5}$

Step3: Calculate $4^5$

$3x+4 = 1024$

Step4: Isolate x term

$3x = 1024 - 4 = 1020$

Step5: Solve for x

$x = \frac{1020}{3} = 340$

Part C

Step1: Combine logs using product rule

$\log_{12}[x(x+1)] = 1$

Step2: Rewrite in exponential form

$x(x+1) = 12^1 = 12$

Step3: Expand and rearrange to quadratic

$x^2 + x - 12 = 0$

Step4: Factor quadratic equation

$(x+4)(x-3) = 0$

Step5: Solve for x, discard invalid solution

$x = -4$ (invalid, log argument can't be negative) or $x = 3$

Answer:

Part A: $x = \log_{12}400 + 1$ or $x \approx 3.41$
Part B: $x = 340$
Part C: $x = 3$