# Exponential equations

• Oct 27th 2007, 05:56 PM
soly_sol
Exponential equations
Sorry for the repost, but I missed up when I typed the problem in.

Solve the exponential equation. Express the solution set in terms of natural logarithms.

4^(x + 4) = 52^(x + 5 )

Solve the exponential equation. Use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution.

7^x = 6^(x + 7 )
• Oct 27th 2007, 06:17 PM
Truthbetold
Use the power (power is in exponent, not as in power-up) rule.
I believe that is the proper name.

(x + 4)^4 = (x + 5 )^ 52 -huh. LaTex doesn't do exponents that big.
$(x + 4) ln 4 = (x + 5) ln 4$

$\frac {(x+4) ln 4}{(x+ 5) ln 52} = 0$ .... or this flipped. It's the same thing I believe.

I believe that is what they are looking for.
I would want a second opinion if I were you.
• Oct 27th 2007, 09:10 PM
Soroban
Hello, soly_sol!

Quote:

Solve the exponential equation. Express the solution set in terms of natural logarithms.

$4^{x + 4} \:= \:52^{x + 5}$

Take logs: . $\ln\left(4^{x+4}\right) \:=\:\ln\left(52^{x+5}\right)\quad\Rightarrow\quad (x+4)\ln(4) \:=\:(x+5)\ln(52)$

Expand: . $x\!\cdot\!\ln(4) + 4\!\cdot\!\ln(4)\:=\:x\!\cdot\!\ln(52) + 5\!\cdot\!\ln(52)$

Rearrange terms: . $x!\cdot\!\ln(4) - x\!\cdot\!\ln(5) \:=\:5\!\cdot\ln(52) - 4\!\cdot\!\ln(4)$

Factor: . $x\left[\ln(4)-\ln(5)\right] \:=\:5\!\cdot\ln(52) - 4\!\cdot\!\ln(4)$

Therefore: . $x \;=\;\frac{5\!\cdot\!\ln(52) - 4\!\cdot\!\ln(4)}{\ln(4) - \ln(5)}$

Quote:

Solve the exponential equation.
Use a calculator to obtain a decimal approximation, to two decimal places.

$7^x \:= \:6^{x + 7}$

Take logs: . $\ln\left(7^x\right)\:=\:\ln\left(6^{x+7}\right) \quad\Rightarrow\quad x\!\cdot\!\ln(7) \:=\:(x+7)\!\cdot\!\ln(6)$

Expand: . $x\!\cdot\!\ln(7) \:=\:x\!\cdot\!\ln(6) + 7\!\cdot\!\ln(6)$

Rearrange: . $x\!\cdot\!\ln(7) - x\!\cdot\!\ln(6)\:=\:7\!\cdot\!\ln(6)$

Factor: . $x\left[\ln(7) - \ln(6)\right] \:=\:7\!\cdot\!\ln(6)$

Therefore: . $x \;=\;\frac{7\!\cdot\!\ln(6)}{\ln(7)-\ln(6)} \;\approx\; 81.36$