I need help with simple algebra with logarithms

Question

Can anyone help me with this? $$xe^{ht} = y $$ How do I eliminate "t" from the LHS of the equation? What I want to end up with is the LHS being a function of "h".

Answer 1

You have that $$e^{ht}=\frac yx,$$ or that $$ht=\log\left(\frac yx\right).$$ Can you continue from here?

Answer 2

$$xe^{ht}=y$$ $$ln(xe^{ht})=ln(y)$$ $$ln(x)+ht=ln(y)$$ $$ht=ln(y)-ln(x)=ln(\frac{y}{x})$$ $$h=\frac{ln(\frac{y}{x})}{t}$$

Answer 3

$$xe^{ht}=x(e^h)^t=y$$

$$(e^h)^t=\frac{y}{x}$$

$$e^h=\bigg(\frac{y}{x}\bigg)^{\frac{1}{t}}$$

Provided that $x,t\neq 0$.