Question

I WILL AWARD MEDAL!!!!!!!!!!!!!!
PLEASE HELP!!!!!!!!!!!!!!!!
Solve 4^(x + 7)= 6^(x – 1)

x ≈ 0.04

x ≈ 0.05

x ≈ 19.51

x ≈ 28.35

Answer 1

@tcarroll010 @whpalmer4 @satellite73

Answer 2

@Hero

Answer 3

x ≈ 19.51

Answer 4

\[x=\frac{-\text{Log}[3]-\text{Log}[32768]}{\text{Log}[2]-\text{Log}[3]}=28.3522 \]

Answer 5

@UnixEditz you are wrong. I got the question wrong with your answer.

Answer 6

Im Sorry Bro..I Meant To Copy and Paste "x ≈ 28.35".Very Sorry.

Answer 7

When x is replaced by 28.3522 in the problem expression, the numerical values of the LHS and RHS is shown below respectively.\[\{1.92373\times 10^{21},1.92375\times 10^{21}\} \]

Answer 8

Probably won't have Mathematica handy on the test, so here's how you work exactly by hand:

\[4^{x+7} = 6^{x-1}\]Rewrite both bases as products
\[(2*2)^{x+7} = (2*3)^{x-1}\]\[2^{2(x+7)} = 2^{x-1}*3^{x-1}\]Divide both sides by the left hand side
\[1 = \frac{2^{x-1}3^{x-1}}{2^{2(x+7)}} = 2^{x-1-2(x+7)}3^{x-1} = 2^{-x-15}3^{x-1}\]Take the log of both sides\[0 = \log(2^{-x-15}3^{x-1}) \]\[\log(a*b) = \log(a)+\log(b) \text{ and }\log(a^n) = n\log(a) \text{, so split up right hand side}\]\[0= \log(2^{-x-15}) + \log(3^{x-1}) = (-x-15)\log (2) + (x-1)\log (3)\]Solve for \(x\)\[0 = -x \log(2) -15 \log(2) + x \log(3) - \log (3)\]\[x(\log(3)-\log(2)) =\log(3) + 15 \log(2)\] \[x = \frac{\log(3) + 15\log(2)}{\log(3)-\log(2)} \approx \frac{0.47712+15*0.30103}{0.47712-0.30103} = 28.35 \]

in case you're wondering about the Log[32768] in Rob's expression, \(2^{15} = 32768\) so \[\log(32768) = \log(2^{15}) = 15\log(2)\]

Answer 9

Even simpler would be instead of dividing by the left half, take log of both sides giving
\[2(x+7)\log (2) = (x-1)\log(2) + (x-1)\log(3)\]then expand and solve for \(x\)