Question

What is the solution to 3^4x-7= 4^2x+3

Answer 1

Is it \[3^{4}x - 7 = 4 ^{2}x + 3\]
or
\[3^{4x} - 7 = 4 ^{2x} + 3\]

Answer 2

or \[3^{4x-7}=4^{2x+3}\]

Answer 3

Sam's right

Answer 4

10/65

Answer 5

explanations please?

Answer 6

\[3^{4x-7}=4^{2x+3}\]
\[\frac{3^{4x}}{3^{7}}=4^{2x}\times {4^3}\]
\[\frac{81^x}{2187} = 64\times16^x\]
\[(\frac{81}{16})^x = 64 \times 2187\]
Now takes logs on both sides

Answer 7

\[x = \frac{(\log 64+\log2187)}{\log 81-\log 16}\]

Answer 8

wow thanks!